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.

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A363636 Indices of numbers of the form k^2+1, k >= 0, that can be written as a product of smaller numbers of that same form.

Original entry on oeis.org

0, 3, 7, 13, 17, 18, 21, 31, 38, 43, 47, 57, 68, 73, 91, 99, 111, 117, 123, 132, 133, 157, 183, 211, 241, 242, 253, 255, 268, 273, 293, 302, 307, 313, 322, 327, 343, 381, 413, 421, 438, 443, 463, 487, 507, 515, 553, 557, 577, 593, 601, 651, 693, 697, 703, 707
Offset: 1

Views

Author

Pontus von Brömssen, Jun 19 2023

Keywords

Comments

For the corresponding sequence for numbers of the form k^3+1 instead of k^2+1, the only terms known to me are 0 and 26, with 26^3+1 = (2^3+1)^2*(6^3+1).

Examples

			0 is a term because 0^2+1 = 1 equals the empty product.
3 is a term because 3^2+1 = 10 = 2*5 = (1^2+1)*(2^2+1).
38 is a term because 38^2+1 = 1445 = 5*17*17 = (2^2+1)*(4^2+1)^2. (This is the first term that requires more than two factors.)
		

Crossrefs

Sequences that list those terms (or their indices or some other key) of a given sequence that are products of smaller terms of the same sequence (in other words, the nonprimitive terms of the multiplicative closure of the sequence):
this sequence (A002522),

Programs

  • Mathematica
    g[lst_, p_] :=
      Module[{t, i, j},
       Union[Flatten[Table[t = lst[[i]]; t[[j]] = p*t[[j]];
          Sort[t], {i, Length[lst]}, {j, Length[lst[[i]]]}], 1],
        Table[Sort[Append[lst[[i]], p]], {i, Length[lst]}]]];
    multPartition[n_] :=
      Module[{i, j, p, e, lst = {{}}}, {p, e} =
        Transpose[FactorInteger[n]];
       Do[lst = g[lst, p[[i]]], {i, Length[p]}, {j, e[[i]]}]; lst];
    output = Join[{0}, Flatten[Position[Table[
         test = Sqrt[multPartition[n^2 + 1][[2 ;; All]] - 1];
         Count[AllTrue[#, IntegerQ] & /@ test, True] > 0
         , {n, 707}], True]]]
    (* David Trimas, Jul 23 2023 *)

A162148 a(n) = n*(n+1)*(5*n+7)/6.

Original entry on oeis.org

0, 4, 17, 44, 90, 160, 259, 392, 564, 780, 1045, 1364, 1742, 2184, 2695, 3280, 3944, 4692, 5529, 6460, 7490, 8624, 9867, 11224, 12700, 14300, 16029, 17892, 19894, 22040, 24335, 26784, 29392, 32164, 35105, 38220, 41514, 44992, 48659, 52520, 56580
Offset: 0

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Author

Keywords

Comments

Partial sums of A147875.
Equals the fourth right hand column of A175136 for n>=1. - Johannes W. Meijer, May 06 2011
a(n) is the number of triples (w,x,y) havingt all terms in {0,...,n} and x+y>w. - Clark Kimberling, Jun 14 2012

Crossrefs

Programs

Formula

a(n) = A162147(n) + A000217(n).
From Johannes W. Meijer, May 06 2011: (Start)
G.f.: x*(4+x)/(1-x)^4.
a(n) = 4*binomial(n+2,3) + binomial(n+1,3).
a(n) = A091894(3,0)*binomial(n+2,3) + A091894(3,1)*binomial(n+1,3). (End)
a(n) = (n+1)*A000290(n+1) - Sum_{i=1..n+1} A000217(i).
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4), a(0)=0, a(1)=4, a(2)=17, a(3)=44. - Harvey P. Dale, May 20 2014
E.g.f.: x*(24 +27*x +5*x^2)*exp(x)/6. - G. C. Greubel, Mar 31 2021

Extensions

Definition rephrased by R. J. Mathar, Jun 27 2009

A220084 a(n) = (n + 1)*(20*n^2 + 19*n + 6)/6.

Original entry on oeis.org

1, 15, 62, 162, 335, 601, 980, 1492, 2157, 2995, 4026, 5270, 6747, 8477, 10480, 12776, 15385, 18327, 21622, 25290, 29351, 33825, 38732, 44092, 49925, 56251, 63090, 70462, 78387, 86885, 95976, 105680, 116017, 127007, 138670, 151026, 164095, 177897, 192452
Offset: 0

Views

Author

Bruno Berselli, Dec 11 2012

Keywords

Comments

Sequence related to heptagonal pyramidal numbers (A002413) by a(n) = n*A002413(n) - (n-1)*A002413(n-1).
Other sequences of numbers of the form m*P(k,m)-(m-1)*P(k,m-1), where P(k,m) is the m-th k-gonal pyramidal number:
k=3, A002412(m) = m*A000292(m)-(m-1)*A000292(m-1);
k=4, A051662(m) = (m+1)*A000330(m+1)-m*A000330(m);
k=5, A213772(m) = m*A002411(m)-(m-1)*A002411(m-1);
k=6, A213837(m) = m*A002412(m)-(m-1)*A002412(m-1);
k=7, this sequence;
k=8, A130748(m) = m*A002414(m)-(m-1)*A002414(m-1).
Also, first bisection of A212983.
Binomial transform of (1, 14, 33, 20, 0, 0, 0, ...). - Gary W. Adamson, Aug 26 2015

Crossrefs

Programs

  • Magma
    [(n+1)*(20*n^2+19*n+6)/6: n in [0..40]]; // Bruno Berselli, Jun 28 2016
    
  • Magma
    /* By first comment: */  A002413:=func; [n*A002413(n)-(n-1)*A002413(n-1): n in [1..40]];
    
  • Magma
    I:=[1,15,62,162]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
    
  • Mathematica
    Table[(n + 1) (20 n^2 + 19 n + 6)/6, {n, 0, 40}]
    LinearRecurrence[{4,-6,4,-1},{1,15,62,162},40] (* Harvey P. Dale, Dec 23 2012 *)
    CoefficientList[Series[(1 + 11 x + 8 x^2) / (1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2013 *)
  • Maxima
    makelist((n+1)*(20*n^2+19*n+6)/6, n, 0, 20); /* Martin Ettl, Dec 12 2012 */
    
  • PARI
    a(n)=(n+1)*(20*n^2+19*n+6)/6 \\ Charles R Greathouse IV, Oct 07 2015

Formula

G.f.: (1+11*x+8*x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3, a(0)=1, a(1)=15, a(2)=62, a(3)=162. - Harvey P. Dale, Dec 23 2012
a(n) = (n+1)*A000566(n+1) + Sum_{i=0..n} A000566(i). - Bruno Berselli, Dec 18 2013
E.g.f.: exp(x)*(6 + 84*x + 99*x^2 + 20*x^3)/6. - Elmo R. Oliveira, Aug 06 2025

A279217 Expansion of Product_{k>=1} 1/(1 - x^k)^(k*(k+1)*(4*k-1)/6).

Original entry on oeis.org

1, 1, 8, 30, 108, 357, 1205, 3838, 12083, 36896, 110828, 326281, 946086, 2700026, 7602642, 21128513, 58028309, 157588912, 423534324, 1127102360, 2971764946, 7766890826, 20131080168, 51766513279, 132117237595, 334770353022, 842462217948, 2106183375971, 5232414548275, 12920429411759, 31719180847831
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 08 2016

Keywords

Comments

Euler transform of the hexagonal pyramidal numbers (A002412).

Crossrefs

Programs

  • Mathematica
    nmax=30; CoefficientList[Series[Product[1/(1 - x^k)^(k (k + 1)(4 k - 1)/6), {k, 1, nmax}], {x, 0, nmax}], x]

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^(k*(k+1)*(4*k-1)/6).
a(n) ~ exp(-Zeta'(-1)/6 - Zeta(3)/(8*Pi^2) - Pi^16/(199065600000*Zeta(5)^3) - Pi^8*Zeta(3)/(6912000*Zeta(5)^2) - Zeta(3)^2/(1440*Zeta(5)) + 2*Zeta'(-3)/3 + (Pi^12/(172800000*2^(4/5)*Zeta(5)^(11/5)) + Pi^4*Zeta(3)/(7200*2^(4/5) * Zeta(5)^(6/5))) * n^(1/5) + (-Pi^8/(288000*2^(3/5)*Zeta(5)^(7/5)) - Zeta(3)/(12*2^(3/5)*Zeta(5)^(2/5))) * n^(2/5) + (Pi^4/(360*2^(2/5)*Zeta(5)^(3/5))) * n^(3/5) + 5*(Zeta(5)/2)^(1/5)/2 * n^(4/5)) * Zeta(5)^(173/1800) / (2^(26/225) * sqrt(5*Pi) * n^(1073/1800)). - Vaclav Kotesovec, Dec 08 2016

A018210 Alkane (or paraffin) numbers l(9,n).

Original entry on oeis.org

1, 4, 16, 44, 110, 236, 472, 868, 1519, 2520, 4032, 6216, 9324, 13608, 19440, 27192, 37389, 50556, 67408, 88660, 115258, 148148, 188552, 237692, 297115, 368368, 453376, 554064, 672792, 811920, 974304, 1162800, 1380825, 1631796
Offset: 0

Views

Author

N. J. A. Sloane, Winston C. Yang (yang(AT)math.wisc.edu)

Keywords

Comments

From M. F. Hasler, May 02 2009: (Start)
Also, 6th column of A159916, i.e., number of 6-element subsets of {1,...,n+6} whose elements add up to an odd integer.
Third differences are A002412([n/2]). (End)
F(1,6,n) is the number of bracelets with 1 blue, 6 identical red and n identical black beads. If F(1,6,1) = 4 and F(1,6,2) = 16 taken as a base, F(1,6,n) = n(n+1)(n+2)(n+3)(n+4)/120 + F(1,4,n) + F(1,6,n-2). F(1,4,n) is the number of bracelets with 1 blue, 4 identical red and n identical black beads. If F(1,4,1) = 3 and F(1,4,2) = 9 taken as a base; F(1,4,n) = n(n+1)(n+2)/6 + F(1,2,n) + F(1,4,n-2). F(1,2,n) is the number of bracelets with 1 blue, 2 identical red and n identical black beads. If F(1,2,1) = 2 and F(1,2,2) = 4 taken as a base F(1,2,n) = n + 1 + F(1,2,n-2). - Ata Aydin Uslu and Hamdi G. Ozmenekse, Mar 16 2012

References

  • S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
  • Winston C. Yang (paper in preparation).

Crossrefs

Cf. A005995 (first differences).

Programs

  • Maple
    a:=n-> (Matrix([[1,0$7,3,12]]). Matrix(10, (i,j)-> if (i=j-1) then 1 elif j=1 then [4, -3, -8, 14, 0, -14, 8, 3, -4, 1][i] else 0 fi)^n)[1,1]: seq (a(n), n=0..33); # Alois P. Heinz, Jul 31 2008
  • Mathematica
    CoefficientList[(1+3*x^2)/((1-x)^7*(1+x)^3) + O[x]^34, x] (* Jean-François Alcover, Jun 08 2015 *)
    LinearRecurrence[{4, -3, -8, 14, 0, -14, 8, 3, -4, 1},{1, 4, 16, 44, 110, 236, 472, 868, 1519, 2520},34] (* Ray Chandler, Sep 23 2015 *)
  • PARI
    A018210(n)=(n+2)*(n+4)*(n+6)^2*(n^2+3*n+5)/1440-if(n%2,(n^2+7*n+11)/32) \\ M. F. Hasler, May 02 2009

Formula

G.f.: (1+3*x^2)/(1-x)^4/(1-x^2)^3. - N. J. A. Sloane
l(c, r) = 1/2 C(c+r-3, r) + 1/2 d(c, r), where d(c, r) is C((c + r - 3)/2, r/2) if c is odd and r is even, 0 if c is even and r is odd, C((c + r - 4)/2, r/2) if c is even and r is even, C((c + r - 4)/2, (r - 1)/2) if c is odd and r is odd.
a(2n) = (n+1)(n+2)(n+3)^2(4n^2+6n+5)/90, a(2n-1) = n(n+1)(n+2)(n+3)(4n^2+6n+5)/90. - M. F. Hasler, May 02 2009
a(n) = (1/(2*6!))*(n+2)*(n+4)*(n+6)*((n+1)*(n+3)*(n+5) + 1*3*5) - (1/2)*(1/2^4)*(n^2+7*n+11)*(1/2)*(1-(-1)^n). - Yosu Yurramendi, Jun 23 2013
a(n) = A060099(n)+3*A060099(n-2). - R. J. Mathar, May 08 2020

A213835 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = 4*n-7+4*h, n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 7, 5, 22, 19, 9, 50, 46, 31, 13, 95, 90, 70, 43, 17, 161, 155, 130, 94, 55, 21, 252, 245, 215, 170, 118, 67, 25, 372, 364, 329, 275, 210, 142, 79, 29, 525, 516, 476, 413, 335, 250, 166, 91, 33, 715, 705, 660, 588, 497, 395
Offset: 1

Views

Author

Clark Kimberling, Jul 04 2012

Keywords

Comments

Principal diagonal: A172078.
Antidiagonal sums: A051797.
Row 1, (1,2,3,4,5,...)**(1,5,9,13,...): A002412.
Row 2, (1,2,3,4,5,...)**(5,9,13,17,...): (4*k^3 + 15*k^2 - 11*k)/6.
Row 3, (1,2,3,4,5,...)**(9,13,17,21,...): (4*k^3 + 27*k^2 - 23*k)/6
For a guide to related arrays, see A212500.

Examples

			Northwest corner (the array is read by falling antidiagonals):
1....7....22....50....95
5....19...46....90....155
9....31...70....130...215
13...43...94....170...275
17...55...118...210...335
21...67...142...250...395
		

Crossrefs

Cf. A212500.
Cf. A304659 (first lower diagonal).

Programs

  • Mathematica
    b[n_]:=n;c[n_]:=4n-3;
    t[n_,k_]:=Sum[b[k-i]c[n+i],{i,0,k-1}]
    TableForm[Table[t[n,k],{n,1,10},{k,1,10}]]
    Flatten[Table[t[n-k+1,k],{n,12},{k,n,1,-1}]]
    r[n_]:=Table[t[n,k],{k,1,60}] (* A213835 *)
    Table[t[n,n],{n,1,40}] (* A172078 *)
    s[n_]:=Sum[t[i,n+1-i],{i,1,n}]
    Table[s[n],{n,1,50}] (* A051797 *)

Formula

T(n,k) = 4*T(n,k-1)-6*T(n,k-2)+4*T(n,k-3)-T(n,k-4).
G.f. for row n: f(x)/g(x), where f(x) = x*((4*n-3) + (4*n-7)*x) and g(x) = (1-x)^4.

A261720 Array of pyramidal (3-dimensional figurate numbers) read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 5, 10, 1, 6, 14, 20, 1, 7, 18, 30, 35, 1, 8, 22, 40, 55, 56, 1, 9, 26, 50, 75, 91, 84, 1, 10, 30, 60, 95, 126, 140, 120, 1, 11, 34, 70, 115, 161, 196, 204, 165, 1, 12, 38, 80, 135, 196, 252, 288, 285, 220, 1, 13, 42, 90, 155, 231, 308, 372, 405, 385, 286
Offset: 1

Views

Author

Gary W. Adamson, Aug 29 2015

Keywords

Comments

First few sequences in the array:
1, 4, 10, 20, 35, 56, 84, 120, 165, 220, ... A000292
1, 5, 14, 30, 55, 91, 140, 204, 285, 385, ... A000330
1, 6, 18, 40, 75, 126, 196, 288, 405, 550, ... A002411
1, 7, 22, 50, 95, 161, 252, 372, 525, 715, ... A002412
1, 8, 26, 60, 115, 196, 308, 456, 645, 880, ... A002413
1, 9, 30, 70, 135, 231, 364, 540, 765, 1045, ... A002414
1, 10, 34, 80, 155, 266, 420, 624, 885, 1210, ... A007584
...
The corresponding bases to rows are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ...

Examples

			Row 2: (1, 5, 14, 30, 55, ...) = (1, 4, 10, 20, 35, ...) + (0, 1, 4, 10, 20, 35, ...).
(1, 7, 22, 50, ...) is the binomial transform of (1, 6, 9, 4, 0, 0, 0, ...) 3rd row in Pascal's triangle (1,4) followed by zeros. (1, 7, 22, 50, ...) is the third partial sum of (1, 4, 4, 4, ...).
		

References

  • Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 194.

Crossrefs

Similar to A080851 but without row n=0.

Programs

  • Mathematica
    T[n_,k_]:=k(k+1)((k-1)n+3)/6; Flatten[Table[T[n-k+1,k],{n,11},{k,n}]] (* Stefano Spezia, Aug 15 2024 *)

Formula

T(n,k) = A080851(n,k).
Given: first sequence in the array is A000292: (1, 4, 10, 20, 35, ...) Subsequent rows are generated by adding (0, 1, 4, 10, 20, 35, ...) to the current row.
n-th row is the binomial transform of row 3 in Pascal's triangle (1,n) followed by zeros. Alternatively, begin with (1, 4, 10, 20, ...) being the binomial transform of (1, 3, 3, 1, 0, 0, 0, ...). Add (0, 1, 2, 1, 0, 0, 0, ...) to the latter to obtain the inverse binomial transform of the next row: (1, 5, 14, 30, 55,..); then repeat the operation.
The row starting (1, N, ...) is the 3rd partial sum of (1, (N-3), (N-3), (N-3), ...).
From Stefano Spezia, Aug 15 2024: (Start)
T(n,k) = k*(k + 1)*((k - 1)*n + 3)/6.
G.f. as array: x*y*(1 + x*(y - 1))/((1 - x)^2*(1 - y)^4).
E.g.f. as array: exp(y)*y*(exp(x)*(6 + 3*(1 + x)*y + x*y^2) - 3*(2 + y))/6. (End)

A279222 Expansion of Product_{k>=1} 1/(1 - x^(k*(k+1)*(4*k-1)/6)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 6, 7, 7, 7, 7, 7, 7, 8, 9, 9, 9, 9, 9, 9, 10, 11, 12, 12, 12, 12, 12, 13, 15, 16, 16, 16, 16, 16, 17, 19, 20, 20, 20, 20, 20, 21, 23, 24, 25, 25, 25, 25, 26, 28, 30, 31, 31, 31, 31, 32, 34, 36, 37, 37, 37, 37, 38, 40, 42, 43, 44, 44, 44
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 08 2016

Keywords

Comments

Number of partitions of n into nonzero hexagonal pyramidal numbers (A002412).

Examples

			a(8) = 2 because we have [7, 1] and [1, 1, 1, 1, 1, 1, 1, 1].
		

Crossrefs

Programs

  • Mathematica
    nmax=90; CoefficientList[Series[Product[1/(1 - x^(k (k + 1) (4 k - 1)/6)), {k, 1, nmax}], {x, 0, nmax}], x]

Formula

G.f.: Product_{k>=1} 1/(1 - x^(k*(k+1)*(4*k-1)/6)).

A105020 Array read by antidiagonals: row n (n >= 0) contains the numbers m^2 - n^2, m >= n+1.

Original entry on oeis.org

1, 3, 4, 5, 8, 9, 7, 12, 15, 16, 9, 16, 21, 24, 25, 11, 20, 27, 32, 35, 36, 13, 24, 33, 40, 45, 48, 49, 15, 28, 39, 48, 55, 60, 63, 64, 17, 32, 45, 56, 65, 72, 77, 80, 81, 19, 36, 51, 64, 75, 84, 91, 96, 99, 100, 21, 40, 57, 72, 85, 96, 105, 112, 117, 120, 121
Offset: 0

Views

Author

Keywords

Comments

A "Goldbach Conjecture" for this sequence: when there are n terms between consecutive odd integers (2n+1) and (2n+3) for n > 0, at least one will be the product of 2 primes (not necessarily distinct). Example: n=3 for consecutive odd integers a(7)=7 and a(11)=9 and of the 3 sequence entries a(8)=12, a(9)=15 and a(10)=16 between them, one is the product of 2 primes a(9)=15=3*5. - Michael Hiebl, Jul 15 2007
A024352 gives distinct values in the array, minus the first row (1, 4, 9, 16, etc.). a(n) gives all solutions to the equation x^2 + xy = n, with y mod 2 = 0, x > 0, y >= 0. - Andrew S. Plewe, Oct 19 2007
Alternatively, triangular sequence of coefficients of Dynkin diagram weights for the Cartan groups C_n: t(n,m) = m*(2*n - m). Row sums are A002412. - Roger L. Bagula, Aug 05 2008

Examples

			Array begins:
  1  4  9 16 25 36  49  64  81 100 ...
  3  8 15 24 35 48  63  80  99 120 ...
  5 12 21 32 45 60  77  96 117 140 ...
  7 16 27 40 55 72  91 112 135 160 ...
  9 20 33 48 65 84 105 128 153 180 ...
  ...
Triangle begins:
   1;
   3,  4;
   5,  8,  9;
   7, 12, 15, 16;
   9, 16, 21, 24, 25;
  11, 20, 27, 32, 35, 36;
  13, 24, 33, 40, 45, 48, 49;
  15, 28, 39, 48, 55, 60, 63, 64;
  17, 32, 45, 56, 65, 72, 77, 80, 81;
  19, 36, 51, 64, 75, 84, 91, 96, 99, 100;
		

References

  • R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Crossrefs

Programs

  • Magma
    [(k+1)*(2*n-k+1): k in [0..n], n in [0..15]]; // G. C. Greubel, Mar 15 2023
    
  • Mathematica
    t[n_, m_]:= (n^2 - m^2); Flatten[Table[t[i, j], {i,12}, {j,i-1,0,-1}]]
    (* to view table *) Table[t[i, j], {j,0,6}, {i,j+1,10}]//TableForm (* Robert G. Wilson v, Jul 11 2005 *)
    Table[(k+1)*(2*n-k+1), {n,0,15}, {k,0,n}]//Flatten (* Roger L. Bagula, Aug 05 2008 *)
  • SageMath
    def A105020(n,k): return (k+1)*(2*n-k+1)
    flatten([[A105020(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Mar 15 2023

Formula

a(n) = r^2 - (r^2 + r - m)^2/4, where r = round(sqrt(m)) and m = 2*n+2. - Wesley Ivan Hurt, Sep 04 2021
a(n) = A128076(n+1) * A105020(n+1). - Wesley Ivan Hurt, Jan 07 2022
From G. C. Greubel, Mar 15 2023: (Start)
Sum_{k=0..n} T(n, k) = A002412(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*((1+(-1)^n)*A000384((n+2)/2) - (1- (-1)^n)*A000384((n+1)/2)). (End)

Extensions

More terms from Robert G. Wilson v, Jul 11 2005

A125233 Triangle T(n,k) read by rows, the (n-k)-th term of the k times repeated partial sum of the hexagonal numbers, 0 <= k < n, 0 < n.

Original entry on oeis.org

1, 6, 1, 15, 7, 1, 28, 22, 8, 1, 45, 50, 30, 9, 1, 66, 95, 80, 39, 10, 1, 91, 161, 175, 119, 49, 11, 1, 120, 252, 336, 294, 168, 60, 12, 1, 153, 372, 588, 630, 462, 228, 72, 13, 1, 190, 525, 960, 1218, 1092, 690, 300, 85, 14, 1, 231, 715, 1485, 2178, 2310, 1782, 990, 385, 99, 15, 1
Offset: 0

Views

Author

Gary W. Adamson, Nov 24 2006

Keywords

Comments

Left border = A000384, hexagonal numbers. The following columns are A002412, A002417, A034263, A051947, ...
Row sums = (1, 7, 23, 59, 135, 291, ...) = A126284.
A125232 is the analogous triangle for the pentagonal numbers.

Examples

			First few rows of the triangle:
   1;
   6,   1;
  15,   7,   1;
  28,  22,   8,   1;
  45,  50,  30,   9,  1;
  66,  95,  80,  39, 10,  1;
  91, 161, 175, 119, 49, 11, 1;
  ...
Example: (5,3) = 80 = 30 + 50 = (4,3) + (4,2).
		

References

  • Albert H. Beiler, "Recreations in the Theory of Numbers", Dover, 1964, p. 189.

Crossrefs

Programs

  • Maple
    A000384Psum:= proc(n,k) coeftayl( x*(1+3*x)/(1-x)^(3+k),x=0,n) ; end: A125233 := proc(n,k) A000384Psum(n-k,k) ; end: for n from 1 to 15 do for k from 0 to n -1 do printf("%d,",A125233(n,k)) ; od: od: # R. J. Mathar, May 03 2008
  • Mathematica
    T[n_, k_] := T[n, k] = Which[k == 0, n (2 n - 1), 1 <= k < n, T[n - 1, k] + T[n - 1, k - 1], True, 0];
    Table[T[n, k], {n, 1, 11}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Sep 14 2023, after R. J. Mathar *)

Formula

T(n,0)=A000384(n). T(n,k) = T(n-1,k) + T(n-1,k-1), k>1. - R. J. Mathar, May 03 2008

Extensions

Edited and extended by R. J. Mathar, May 03 2008, and M. F. Hasler, Sep 29 2012
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