cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A086787 a(n) = Sum_{i=1..n} ( Sum_{j=1..n} i^j ).

Original entry on oeis.org

1, 8, 56, 494, 5699, 82200, 1419760, 28501116, 651233661, 16676686696, 472883843992, 14705395791306, 497538872883727, 18193397941038736, 714950006521386976, 30046260016074301944, 1344648068888240941017
Offset: 1

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Aug 04 2003

Keywords

Comments

p divides a(p+1) for all prime p except 3. p^2 divides a(p+1) for prime p in A123374.
2 divides a(n) for n = {2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 15, 16, 18, 19, 20, 22, 23, 24, 26, 27, 28, 30, 31, 32, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, ...}.
2^2 divides a(n) for n = {2, 3, 6, 7, 8, 10, 11, 14, 15, 16, 18, 19, 22, 23, 24, 26, 27, 30, 31, 32, 34, 35, 38, 39, 40, 42, 43, 46, 47, 48, 50, ...}.
2^3 divides a(n) for n = {2, 3, 6, 7, 10, 11, 14, 15, 16, 18, 19, 22, 23, 26, 27, 30, 31, 32, 34, 35, 38, 39, 42, 43, 46, 47, 48, 50, ...}.
2^4 divides a(n) for n = {7, 14, 15, 18, 23, 30, 31, 32, 34, 39, 46, 47, 50, ...}.
2^5 divides a(n) for n = {15, 30, 31, 34, 47, 62, 63, 64, 66, 79, 94, 95, 98, ...}.
2^6 divides a(n) for n = {31, 62, 63, 66, 95, ...}.
2^7 divides a(n) for n = {63, 126, 127, 130, ...}.
It appears that for k > 2 the least few n such that a(n) is divisible by 2^(k+1) are n = {(2^k-1), 2*(2^k-1), 2*(2^k-1)+1, 2*(2^k-1)+3, 3*(2^k-1)+2, 4*(2^k-1)+2, 4*(2^k-1)+3, 4*(2^k-1)+4, 4*(2^k-1)+6, 5*(2^k-1)+4, 6*(2^k-1)+4, 6*(2^k-1)+5, 6*(2^k-1)+8, ...}. - Alexander Adamchuk, Oct 08 2006
Numbers n that divide a(n) are listed in A014741. - Alexander Adamchuk, Nov 03 2006

Examples

			a(2) = 8 = 1 + 1 + 2 + 4 = 1^1 + 1^2 + 2^1 + 2^2.

Links

Crossrefs

Cf. A000295, A014741, A215084, A349117.

Programs

  • Maple
    seq(1-Psi(n)-gamma+sum(i^(n+1)/(i-1),i = 2 .. n),n=1..20);
  • Mathematica
    Table[Sum[i^j,{i,1,n},{j,1,n}],{n,1,24}] (* Alexander Adamchuk, Oct 08 2006 *)
Table[ n + Sum[ i*(i^n-1)/(i-1), {i,2,n} ], {n,1,17} ] (* Alexander Adamchuk, Nov 03 2006 *)
  • PARI
    a(n)=sum(i=1,n,sum(j=1,n,i^j)) \\ Charles R Greathouse IV, Jul 19 2013
  • PARI
    a(n)=round(1-psi(n)-Euler+sum(i=2,n,i^(n+1)/(i-1))) \\ Charles R Greathouse IV, Jul 19 2013
  • Python
    def A086787(n): return sum(i**j for i in range(1,n+1) for j in range(1,n+1)) # Chai Wah Wu, Jan 08 2022
  • Python
    from sympy import digamma, EulerGamma
from fractions import Fraction
def A086787(n): return 1-digamma(n)-EulerGamma + sum(Fraction(i**(n+1),i-1) for i in range(2,n+1)) # Chai Wah Wu, Jan 08 2022

Formula

1 - Psi(n) - gamma + Sum_{i=2..n} (i^(n+1)/(i-1)), where Psi(n) is the digamma function and gamma is Euler's constant.
a(n) = Sum[ i^j, {i,1,n}, {j,1,n} ] = n + Sum[ i*(i^n - 1)/(i - 1), {i,2,n} ]. - Alexander Adamchuk, Nov 03 2006
a(n) = Sum_{k=1..n} (B(k+1, n+1) - B(k+1, 1))/(k+1), where B(n, x) are the Bernoulli polynomials. - Daniel Suteu, Jun 25 2018
a(n) ~ c * n^n, where c = 1 / (1 - exp(-1)) = A185393 = 1.58197670686932642438... - Vaclav Kotesovec, Nov 06 2021

Extensions

Edited by Max Alekseyev, Jan 29 2012

A215077 Binomial convolution of sum of consecutive powers.

Original entry on oeis.org

0, 1, 7, 66, 852, 14020, 280472, 6609232, 179317056, 5505532992, 188717617280, 7143999854464, 296013377405440, 13325516967972352, 647610246703508480, 33794224057227356160, 1884620857353101983744, 111857608180484932648960, 7040178644779119413723136, 468349192560992552808841216, 32836927387372039917034405888
Offset: 0

Views

Author

Olivier Gérard, Aug 02 2012

Keywords

Comments

a(0) could alternatively be defined as 1 from the formula or the convention for 0^0.
This sum is remarkable for its three different decompositions involving powers and binomials (see formulas and cross-refs).

Links

Crossrefs

Row sums of A215078, A215079, A215080.
See also A215083 and A215084.

Programs

  • Mathematica
    Table[Sum[Sum[j^n*Binomial[n, k], {j, 1, k}], {k, 0, n}], {n, 0, 20}]
  • PARI
    a(n)=sum(k=0,n, binomial(n,k)*sum(j=1,k, j^n)) \\ Charles R Greathouse IV, Jul 31 2016
  • PARI
    a(n)=my(P=sumformal('x^n)); sum(k=0,n, binomial(n,k)*subst(P,'x,k)) \\ Charles R Greathouse IV, Jul 31 2016

Formula

a(n) = Sum_{k=0..n} binomial(n,k)*Sum_{j=1..k} j^n;
a(n) = Sum_{k=0..n} binomial(n,k)*H_k^{-n}, where H_k^(-n) = k-th harmonic number of order -n;
a(n) = Sum_{k=0..n} k^n * Sum_{j=0..n-k} binomial(n,n-k-j);
a(n) = Sum_{k=0..n} k^n * binomial(n,n-k) * 2F1(1, k-n; k+1)(-1);
a(n) = Sum_{k=0..n} Sum_{j=0..k} (k-j)^n * binomial(n,j);
a(n) = Sum_{k=0..n} Sum_{j=0..n} (n-j)^n * binomial(n,n+k-j);
and the equivalent formulas obtained by symmetries of the binomial and the hypergeometric function as well as treating the zeroth term separately.
a(n) ~ n^n / (sqrt(1+r) * (1-r) * exp(n) * r^n), where r = A202357 = LambertW(exp(-1)). - Vaclav Kotesovec, Jun 10 2019

A349116 a(n) = Sum_{m=1..n} (Sum_{k=1..m} (Sum_{j=1..k} j^n)).

Original entry on oeis.org

1, 7, 57, 605, 7990, 126378, 2331462, 49183590, 1168343775, 30871159429, 898302280271, 28547663567787, 983905660878140, 36556999252315556, 1456715662183096092, 61973464872507939132, 2803744852868605501965, 134413716665674685292183, 6807005707226954237422477
Offset: 1

Views

Author

Vaclav Kotesovec, Nov 08 2021

Keywords

Links

Crossrefs

Cf. A215084, A349117.

Programs

  • Mathematica
    Table[Sum[Sum[Sum[j^n, {j, 1, k}], {k, 1, m}], {m, 1, n}], {n, 1, 20}]

Formula

a(n) ~ c * n^n, where c = 1/(1 - 3/exp(1) + 3/exp(2) - 1/exp(3)) = 3.959134481...
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.