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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A001296 4-dimensional pyramidal numbers: a(n) = (3*n+1)*binomial(n+2, 3)/4. Also Stirling2(n+2, n).

Original entry on oeis.org

0, 1, 7, 25, 65, 140, 266, 462, 750, 1155, 1705, 2431, 3367, 4550, 6020, 7820, 9996, 12597, 15675, 19285, 23485, 28336, 33902, 40250, 47450, 55575, 64701, 74907, 86275, 98890, 112840, 128216, 145112, 163625, 183855, 205905, 229881, 255892, 284050, 314470
Offset: 0

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Author

Keywords

Comments

Permutations avoiding 12-3 that contain the pattern 31-2 exactly once.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
Partial sums of A002411. - Jonathan Vos Post, Mar 16 2006
If Y is a 3-subset of an n-set X then, for n>=6, a(n-5) is the number of 6-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
Starting with 1 = binomial transform of [1, 6, 12, 10, 3, 0, 0, 0, ...]. Equals row sums of triangle A143037. - Gary W. Adamson, Jul 18 2008
Rephrasing the Perry formula of 2003: a(n) is the sum of all products of all two numbers less than or equal to n, including the squares. Example: for n=3 the sum of these products is 1*1 + 1*2 + 1*3 + 2*2 + 2*3 + 3*3 = 25. - J. M. Bergot, Jul 16 2011
Half of the partial sums of A011379. [Jolley, Summation of Series, Dover (1961), page 12 eq (66).] - R. J. Mathar, Oct 03 2011
Also the number of (w,x,y,z) with all terms in {1,...,n+1} and w < x >= y > z (see A211795). - Clark Kimberling, May 19 2012
Convolution of A000027 with A000326. - Bruno Berselli, Dec 06 2012
This sequence is related to A000292 by a(n) = n*A000292(n) - Sum_{i=0..n-1} A000292(i) for n>0. - Bruno Berselli, Nov 23 2017
a(n-2) is the maximum number of intersections made from the perpendicular bisectors of all pair combinations of n points. - Ian Tam, Dec 22 2020

Examples

			G.f. = x + 7*x^2 + 25*x^3 + 65*x^4 + 140*x^5 + 266*x^6 + 462*x^7 + 750*x^8 + 1155*x^9 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 227, #16.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 166, Table 10.4/I/3).
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n)=f(n, 2) where f is given in A034261.
a(n)= A093560(n+3, 4), (3, 1)-Pascal column.
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.
Cf. similar sequences listed in A241765 and A254142.
Cf. A000914.

Programs

  • Magma
    /* A000027 convolved with A000326: */ A000326:=func; [&+[(n-i+1)*A000326(i): i in [0..n]]: n in [0..40]]; // Bruno Berselli, Dec 06 2012
    
  • Magma
    [(3*n+1)*Binomial(n+2,3)/4: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
  • Maple
    A001296:=-(1+2*z)/(z-1)**5; # Simon Plouffe in his 1992 dissertation for sequence without the leading zero
  • Mathematica
    Table[n*(1+n)*(2+n)*(1+3*n)/24, {n, 0, 100}]
    CoefficientList[Series[x (1 + 2 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
    Table[StirlingS2[n+2, n], {n, 0, 40}] (* Jean-François Alcover, Jun 24 2015 *)
    Table[ListCorrelate[Accumulate[Range[n]],Range[n]],{n,0,40}]//Flatten (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,7,25,65},40] (* Harvey P. Dale, Aug 14 2017 *)
  • PARI
    t(n)=n*(n+1)/2
    for(i=1,30,print1(","sum(j=1,i,j*t(j))))
    
  • PARI
    {a(n) = n * (n+1) * (n+2) * (3*n+1) / 24}; /* Michael Somos, Sep 04 2017 */
    
  • Sage
    [stirling_number2(n+2,n) for n in range(0,38)] # Zerinvary Lajos, Mar 14 2009
    

Formula

a(n) = n*(1+n)*(2+n)*(1+3*n)/24. - T. D. Noe, Jan 21 2008
G.f.: x*(1+2*x)/(1-x)^5. - Paul Barry, Jul 23 2003
a(n) = Sum_{j=0..n} j*A000217(j). - Jon Perry, Jul 28 2003
E.g.f. with offset -1: exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!). For the coefficients [1, 4, 3] see triangle A112493.
E.g.f. x*exp(x)*(24 + 60*x + 28*x^2 + 3*x^3)/24 (above e.g.f. differentiated).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 3. - Kieren MacMillan, Sep 29 2008
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Jaume Oliver Lafont, Nov 23 2008
O.g.f. is D^2(x/(1-x)) = D^3(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
a(n) = A153978(n)/2. - J. M. Bergot, Aug 09 2013
a(n) = A002817(n) + A000292(n-1). - J. M. Bergot, Aug 29 2013; [corrected by Cyril Damamme, Feb 26 2018]
a(n) = A000914(n+1) - 2 * A000330(n+1). - Antal Pinter, Dec 31 2015
a(n) = A080852(3,n-1). - R. J. Mathar, Jul 28 2016
a(n) = 1*(1+2+...+n) + 2*(2+3+...+n) + ... + n*n. For example, a(6) = 266 = 1(1+2+3+4+5+6) + 2*(2+3+4+5+6) + 3*(3+4+5+6) + 4*(4+5+6) + 5*(5+6) + 6*(6).- J. M. Bergot, Apr 20 2017
a(n) = A000914(-2-n) for all n in Z. - Michael Somos, Sep 04 2017
a(n) = A000292(n) + A050534(n+1). - Cyril Damamme, Feb 26 2018
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = (6/5) * (47 - 3*sqrt(3)*Pi - 27*log(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (6/5) * (16*log(2) + 6*sqrt(3)*Pi - 43). (End)

A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

Original entry on oeis.org

0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422
Offset: 0

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Author

Keywords

Comments

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

Examples

			4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - _Bruno Berselli_, Feb 05 2014
		

References

  • Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

Crossrefs

Programs

  • Haskell
    a024166 n = sum $ zipWith (*) [n+1,n..0] a000578_list
    -- Reinhard Zumkeller, Oct 14 2001
    
  • Magma
    [n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017
    
  • Maple
    A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017
  • Mathematica
    Nest[Accumulate,Range[0,40]^3,2] (* Harvey P. Dale, Jan 10 2016 *)
    Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n,0,30}] (* G. C. Greubel, Nov 21 2017 *)
  • PARI
    a(n)=sum(j=1,n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008
    
  • PARI
    for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

Formula

From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022

A176541 Numbers n such that there exist n consecutive triangular numbers which sum to a square.

Original entry on oeis.org

0, 1, 2, 3, 4, 11, 13, 22, 23, 25, 27, 32, 37, 39, 46, 47, 48, 49, 50, 52, 59, 66, 71, 73, 83, 94, 98, 100, 104, 107, 109, 111, 118, 121, 128, 143, 146, 147, 148, 157, 167, 176, 179, 181, 183, 191, 192, 193, 194, 200, 214, 219, 227, 239, 241, 242, 243, 244, 253, 263
Offset: 1

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Author

Andrew Weimholt, Apr 20 2010

Keywords

Comments

Numbers n such that there exists some x >= 0 such that A000292(x+n) - A000292(x) is a square. Terms of this sequence, for which only a finite number of solutions x exist, are given in A176542.
Integer n is in the sequence if there exist non-degenerate solutions to the Diophantine equation: 8x^2 - n*y^2 - A077415(n) = 0. A degenerate solution is one involving triangular numbers with negative indexes.
The sum of n consecutive triangular numbers starting at the j-th is Sum_{k=j..j+n-1} A000217(k) = n*(n^2 + 3*j*n + 3*j^2 - 1)/6, see A143037. - R. J. Mathar, May 06 2015

Examples

			0 is in the sequence because the sum of 0 consecutive triangular numbers is 0 (a square).
1 is in the sequence because there exist triangular numbers which are squares (cf. A001110).
2 is in the sequence because ANY 2 consecutive triangular numbers sum to a square.
3 is in the sequence because there are infinitely many solutions (cf. A165517).
4 is in the sequence because there infinitely many solutions (cf. A202391).
5 is NOT in the sequence because no 5 consecutive triangular numbers sum to a square.
For n=8, solutions to the Diophantine equation exist, but start at A000217(-2) and A000217(-6): 1 + 0 + 0 + 1 + 3 + 6 + 10 + 15 = 36 and 15 + 10 + 6 + 3 + 1 + 0 + 0 + 1 = 36. There are no non-degenerate solutions for n=8. Hence, 8 is not included in the sequence.
For n=11, there exist infinitely many solutions (cf. A116476), so 11 is in the sequence.
		

Crossrefs

Extensions

More terms from Max Alekseyev, May 10 2010
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