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.

A132318 Triangle, read by rows, where T(n,k) = [x^(k*2^(n-1))] Product_{i=0..n-1} (1 + x^(2^i))^(2^(n-i-1)) for n>0 with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 15, 15, 1, 1, 1024, 2046, 1024, 1, 1, 7048181, 60060682, 60060682, 7048181, 1, 1, 469389728563470, 72057594037927935, 143176408618728932, 72057594037927935, 469389728563470, 1, 1, 2954306864416502250656677496683
Offset: 0

Views

Author

Paul D. Hanna, Aug 19 2007

Keywords

Comments

There are n*2^(n-1)+1 coefficients in P(n) = Product_{i=0..n-1} (1 + x^(2^i))^(2^(n-i-1)) for n>0; in this triangle, row n consists of coefficients of x^(k*2^(n-1)) in P(n) as k=0..n.

Examples

			Triangle begins:
1;
1,1;
1,2,1;
1,15,15,1;
1,1024,2046,1024,1;
1,7048181,60060682,60060682,7048181,1;
1,469389728563470,72057594037927935,143176408618728932,72057594037927935,469389728563470,1;
Examples:
T(2,1) = [x^(1*2)] (1+x)^2*(1+x^2) = 2;
T(3,1) = [x^(1*4)] (1+x)^4*(1+x^2)^2*(1+x^4) = 15;
T(4,3) = [x^(3*8)] (1+x)^8*(1+x^2)^4*(1+x^4)^2*(1+x^8) = 1024;
T(5,3) = [x^(3*16)] (1+x)^16*(1+x^2)^8*(1+x^4)^4*(1+x^8)^2*(1+x^16) = 60060682.
		

Crossrefs

Cf. A132317 (column 1), A132316.

Programs

  • PARI
    {T(n,k)=if(n==0,1,polcoeff(prod(i=0,n-1,(1+x^(2^i)+x*O(x^(k*2^(n-1))))^(2^(n-i-1))),k*2^(n-1)))}

Formula

Row sums equal 2^(2^n - n) for n>0 - improved formula and proof by Max Alekseyev, Aug 19 2007.

A132316 a(n) = [x^n] Product_{i=0..n} (1 + x^(2^i) )^(2^(n-i)).

Original entry on oeis.org

1, 2, 8, 88, 2812, 284832, 96344064, 112162777984, 458279216351168, 6667184111642112512, 349410072608155198029824, 66605152356815910201401874432, 46557942811582437260863430233248768
Offset: 0

Views

Author

Paul D. Hanna, Aug 18 2007

Keywords

Examples

			a(2) = [x^2] (1+x)^4*(1+x^2)^2*(1+x^4) = 8;
a(3) = [x^3] (1+x)^8*(1+x^2)^4*(1+x^4)^2*(1+x^8) = 88;
a(4) = [x^4] (1+x)^16*(1+x^2)^8*(1+x^4)^4*(1+x^8)^2*(1+x^16) = 2812.
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[(1 + x^(2^j))^(2^(n-j)),{j,0,n}],{x,0,n}], {n,0,15}] (* Vaclav Kotesovec, Oct 09 2020 *)
  • PARI
    {a(n)=polcoeff(prod(i=0,#binary(n),(1 + x^(2^i) +x*O(x^n))^(2^(n-i))), n)}

Formula

a(n) ~ 2^(n^2) / n!. - Vaclav Kotesovec, Oct 09 2020

A175631 a(n) = (n-th pentagonal number) modulo (n-th triangular number).

Original entry on oeis.org

0, 2, 0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430
Offset: 1

Views

Author

Zak Seidov, Jul 29 2010

Keywords

Examples

			a(1)=0 because (1(3-1)/2) mod (1(1+1)/2) = 1 mod 1 = 0,
a(2)=2 because (2(6-1)/2) mod (2(2+1)/2) = 5 mod 3 = 2.
		

Crossrefs

Cf. A000096 (n(n+3)/2), A000217 (triangular numbers), A000326 (pentagonal numbers), A175630 (n-th pentagonal number mod (n+2)).

Programs

  • Magma
    [n lt 4 select 1+(-1)^n else n*(n-3)/2: n in [1..60]]; // G. C. Greubel, Jan 30 2022
    
  • Mathematica
    Table[Mod[n(3n-1)/2, n(n+1)/2],{n,100}]
    Module[{nn=60},Mod[#[[1]],#[[2]]]&/@Thread[{PolygonalNumber[ 5,Range[ nn]],Accumulate[ Range[nn]]}]] (* Harvey P. Dale, Nov 19 2022 *)
  • Sage
    def A175631(n): return 1+(-1)^n if (n<4) else 9*binomial(n/3, 2)
    [A175631(n) for n in (1..60)] # G. C. Greubel, Jan 30 2022

Formula

For n>=3, a(n) = A000096(n-2).
From Chai Wah Wu, Oct 12 2018: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 5.
G.f.: x^2*(2 - 6*x + 8*x^2 - 3*x^3)/(1 - x)^3. (End)
E.g.f.: (x/2)*(2 + 3*x - (2 - x)*exp(x)). - G. C. Greubel, Jan 30 2022
Showing 1-3 of 3 results.