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 1-5 of 5 results.

A357671 a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).

Original entry on oeis.org

2, 4, 20, 166, 1812, 22504, 297362, 4067298, 56897300, 809019580, 11649254520, 169444978124, 2485270719570, 36707044807996, 545386321069862, 8144809732228666, 122177690210103060, 1839933274439787940, 27804610626798500372, 421476329345312885304, 6406685025104178888312
Offset: 0

Views

Author

Peter Bala, Oct 10 2022

Keywords

Comments

Conjectures:
1) a(p) == 4 (mod p^5) for all primes p >= 7 (checked up to p = 499). Note that A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5.
2) For r >= 2, and all primes p >= 5, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
3) More generally, let m be a positive integer and set u(n) = 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k) + (m + 1)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2. Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 1. See A357673 for the case m = 2.
4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

Examples

			Examples of supercongruences:
a(19) - a(1) = 421476329345312885304 - 4 = (2^2)*(5^2)*(19^5)*1913*2383*373393 == 0 (mod 19^5).
a(25) - a(5) = 5375188503768910125546897504 - 22504 = (2^3)*(5^10)*1858537* 37019662696111 == 0 (mod 5^10).

Crossrefs

Cf. A000984, A333592, A357509, A357565, A357566, A357672, A357673, A357674.

Programs

  • Maple
    seq(add( binomial(n+k-1,k) + binomial(n+k-1,k)^2, k = 0..n ), n = 0..20);
  • PARI
    a(n) = sum(k = 0, n, binomial(n+k-1,k) + binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022
  • Python
    from math import comb
def A357671(n): return comb(n<<1,n)+sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 2 # Chai Wah Wu, Oct 28 2022

Formula

a(n) = A000984(n) + A333592(n).
a(p) == 4 (mod p^3) for all primes p >= 5.
a(n) ~ 2^(4*n) / (3*Pi*n).

A357565 a(n) = 3*Sum_{k = 0..n} binomial(n+k-1,k)^2 + 2*Sum_{k = 0..n} binomial(n+k-1,k)^3.

Original entry on oeis.org

5, 10, 114, 2926, 109106, 4846260, 234488526, 11913003294, 625130924082, 33590792825200, 1838547540484364, 102135528447552060, 5743779960435245774, 326352202770939600460, 18706076476872783254286, 1080345839256279791104926, 62806507721442655949609010
Offset: 0

Views

Author

Peter Bala, Oct 16 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^5) for all odd primes p except p = 5 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 3.
3) More generally, let m be a positive integer and set u(n) = (m + 2)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2 + 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k)^3. Then the supercongruences u(p) == u(1) (mod p^5) hold for all primes p >= 7.
4) u(p^r) == u(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 3.

Examples

			a(11) - a(1) = 102135528447552060 - 10 = 2*(5^2)*(11^5)*14657* 865363 == 0 (mod 11^5).
a(5^2) - a(5) = 581553752659150682384860284864053981408760 - 4846260 = 3*(2^2)*(5^9)*5611847956825027*4421531072180960789 == 0 (mod 5^9)

Crossrefs

Cf. A357566, A357671, A357672, A357673, A357674.

Programs

  • Maple
    seq(add( 3*binomial(n+k-1,k)^2 + 2*binomial(n+k-1,k)^3, k = 0..n ), n = 0..20);
  • PARI
    a(n) = 3*sum(k = 0, n, binomial(n+k-1,k)^2) + 2*sum(k = 0, n, binomial(n+k-1,k)^3); \\ Michel Marcus, Oct 25 2022

A357566 a(n) = ( Sum_{k = 0..n} binomial(n+k-1,k)^2 )^3 * ( Sum_{k = 0..n} binomial(n+k-1,k)^3 )^2.

Original entry on oeis.org

1, 32, 3556224, 4816142496896, 14260946236464636800, 62923492736113950202540032, 355372959542696519903013302282592, 2376354966106399942850054560101358877184, 17973185649572984869873798116070605084766512000, 149319509846904520286037745483655872001727895961600000
Offset: 0

Views

Author

Peter Bala, Oct 16 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^5) for all odd primes p except p = 5 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.
3) More generally, let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+2) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^3 )^(2*m). Then the supercongruences u(p) == u(1) (mod p^5) hold for all primes p >= 5.
4) u(p^r) == u(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.

Examples

			a(7) - a(1) = 2376354966106399942850054560101358877184 - 32 = (2^5)*(7^5)*19*31*317*339247*25170329*2771351868561767 == 0 (mod 7^5).

Crossrefs

Cf. A357565, A357671, A357672, A357673, A357674.
Cf. A333592.

Programs

  • Maple
    seq((add(binomial(n+k-1,k)^2, k = 0..n))^3 * (add( binomial(n+k-1,k)^3, k = 0..n))^2, n = 0..20);
  • Mathematica
    Table[Sum[Binomial[n+k-1,k]^2, {k, 0, n}]^3 * Sum[Binomial[n+k-1,k]^3, {k, 0, n}]^2, {n, 0, 10}] (* Vaclav Kotesovec, May 31 2025 *)
  • PARI
    a(n) = sum(k = 0, n, binomial(n+k-1,k)^2)^3 * sum(k = 0, n, binomial(n+k-1,k)^3)^2; \\ Michel Marcus, Oct 25 2022

Formula

a(n) ~ 2^(24*n) / (1323*Pi^6*n^6). - Vaclav Kotesovec, May 31 2025

A357672 a(n) = Sum_{k = 0..n} binomial(n+k-1,k) * Sum_{k = 0..n} binomial(n+k-1,k)^2.

Original entry on oeis.org

1, 4, 84, 2920, 121940, 5607504, 273908712, 13947188112, 732102614100, 39332168075200, 2152235533317584, 119531412173662944, 6720552415489860584, 381775182057562837600, 21879043278489630349200, 1263402662473729731877920, 73438613319490294002441300, 4293679728171938162242298400
Offset: 0

Views

Author

Peter Bala, Oct 10 2022

Keywords

Comments

Conjectures:
1) a(p) == 4 (mod p^5) for all odd primes p except p = 5 (checked up to p = 499). Note that both A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5 and hence a(p) == 4 (mod p^3) for all primes p >= 5.
2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
3) More generally, let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k) )^(2*m) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+1). Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. See A357674 for the case m = 2.
4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

Examples

			Examples of supercongruences:
a(17) - a(1) = 4293679728171938162242298400 - 4 = (2^2)*(17^5)*3457* 218688360593678551 == 0 (mod 17^5).
a(5^2) - a(5) = (2^4)*(3^2)*(5^9)*7*7229*102559*465516030080883405648119 == 0 (mod 5^9).

Crossrefs

Cf. A000984, A333592, A357565, A357566, A357671, A357673, A357674.

Programs

  • Maple
    seq(add(binomial(n+k-1,k), k = 0..n) * add( binomial(n+k-1,k)^2, k = 0..n), n = 0..20);
  • PARI
    a(n) = sum(k = 0, n, binomial(n+k-1,k)) * sum(k = 0, n, binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022

Formula

a(n) = A000984(n) * A333592(n).
a(n) ~ 2^(6*n)/(3*(Pi*n)^(3/2)).

A357674 a(n) = ( Sum_{k = 0..2*n} binomial(n+k-1,k) )^4 * ( Sum_{k = 0..2*n} binomial(n+k-1,k)^2 )^3.

Original entry on oeis.org

1, 2187, 8422734375, 202402468703748096, 9223976224194016590174375, 587835594121137662072707812564687, 46157429480574073282465608886521546620928, 4181198339699286332943143923058721957212160000000, 420336565507755143573799144638372909582306681004894518439
Offset: 0

Views

Author

Peter Bala, Oct 11 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 271).
2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
3) Let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k) )^(2*m) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+1). Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 2. See A357672 for the case m = 1.

Examples

			Example of a supercongruence:
a(7) - a(1) = 4181198339699286332943143923058721957212160000000 - 2187 = (3^7)*(7^5)*211*298225180113209*1807736060307048120859243 == 0 (mod 7^5).

Crossrefs

Cf. A005809, A357565, A357566, A357671, A357672, A357673.

Programs

  • Maple
    seq((add(binomial(n+k-1,k), k = 0..2*n))^4 * (add( binomial(n+k-1,k)^2, k = 0..2*n))^3, n = 0..20);
  • Mathematica
    Table[Binomial[3*n,n]^4 * Sum[Binomial[n+k-1,k]^2, {k, 0, 2*n}]^3, {n, 0, 10}] (* Vaclav Kotesovec, May 31 2025 *)
  • PARI
    a(n) = sum(k = 0, 2*n, binomial(n+k-1,k))^4 * sum(k = 0, 2*n, binomial(n+k-1,k)^2)^3; \\ Michel Marcus, Oct 24 2022

Formula

a(n) = ( A005809(n) )^4 * (Sum_{k = 0..2*n} binomial(n+k-1,k)^2 )^3.
a(n) ~ 3^(30*n+5) / (125 * Pi^5 * n^5 * 2^(20*n+10)). - Vaclav Kotesovec, May 31 2025
Showing 1-5 of 5 results.

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