Very small constants from poly-bernoulli
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If we define
$$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
$$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
for $n>0$, $mgeqslant0$, so
$$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
for $n>0$, $|m|geqslant0$, ex.
$$a_1(m)-2a_1(m-1)=0$$
$$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
Another words
$$a_1(m)=fraca_1(m+1)2$$
$$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
Then if
$$f(n)=sumlimits_k=1^inftya_n(-k)$$
so $f(1)=1$, $f(2)=0$.
For $n>2$ we have very small constants with rational ratio, ex.
$$g(n)=fracf(n)f(3)$$
$$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
$$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
Also, surprisingly, it looks like $f(3)=2^-55$.
Is there a way to identify them without precisely calculation?
calculus recurrence-relations interpolation constants bernoulli-numbers
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up vote
0
down vote
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If we define
$$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
$$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
for $n>0$, $mgeqslant0$, so
$$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
for $n>0$, $|m|geqslant0$, ex.
$$a_1(m)-2a_1(m-1)=0$$
$$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
Another words
$$a_1(m)=fraca_1(m+1)2$$
$$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
Then if
$$f(n)=sumlimits_k=1^inftya_n(-k)$$
so $f(1)=1$, $f(2)=0$.
For $n>2$ we have very small constants with rational ratio, ex.
$$g(n)=fracf(n)f(3)$$
$$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
$$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
Also, surprisingly, it looks like $f(3)=2^-55$.
Is there a way to identify them without precisely calculation?
calculus recurrence-relations interpolation constants bernoulli-numbers
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we define
$$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
$$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
for $n>0$, $mgeqslant0$, so
$$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
for $n>0$, $|m|geqslant0$, ex.
$$a_1(m)-2a_1(m-1)=0$$
$$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
Another words
$$a_1(m)=fraca_1(m+1)2$$
$$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
Then if
$$f(n)=sumlimits_k=1^inftya_n(-k)$$
so $f(1)=1$, $f(2)=0$.
For $n>2$ we have very small constants with rational ratio, ex.
$$g(n)=fracf(n)f(3)$$
$$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
$$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
Also, surprisingly, it looks like $f(3)=2^-55$.
Is there a way to identify them without precisely calculation?
calculus recurrence-relations interpolation constants bernoulli-numbers
If we define
$$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
$$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
for $n>0$, $mgeqslant0$, so
$$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
for $n>0$, $|m|geqslant0$, ex.
$$a_1(m)-2a_1(m-1)=0$$
$$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
Another words
$$a_1(m)=fraca_1(m+1)2$$
$$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
Then if
$$f(n)=sumlimits_k=1^inftya_n(-k)$$
so $f(1)=1$, $f(2)=0$.
For $n>2$ we have very small constants with rational ratio, ex.
$$g(n)=fracf(n)f(3)$$
$$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
$$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
Also, surprisingly, it looks like $f(3)=2^-55$.
Is there a way to identify them without precisely calculation?
calculus recurrence-relations interpolation constants bernoulli-numbers
calculus recurrence-relations interpolation constants bernoulli-numbers
edited Sep 4 at 10:41
asked Sep 4 at 9:09
user565184
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