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 1-10 of 19 results. Next

A051843 Partial sums of A002419.

Original entry on oeis.org

0, 1, 11, 51, 161, 406, 882, 1722, 3102, 5247, 8437, 13013, 19383, 28028, 39508, 54468, 73644, 97869, 128079, 165319, 210749, 265650, 331430, 409630, 501930, 610155, 736281, 882441, 1050931, 1244216, 1464936, 1715912, 2000152, 2320857, 2681427
Offset: 0

Views

Author

Barry E. Williams, Dec 13 1999

Keywords

Comments

5-dimensional form of octagonal-based pyramidal numbers. - Derek I. Thomas (dithom02(AT)louisville.edu), Jun 30 2007
Convolution of triangular numbers (A000217) and octagonal numbers (A000567). [Bruno Berselli, Jul 21 2015]
Also the number of 4-cycles in the (n+2)-triangular honeycomb bishop graph. - Eric W. Weisstein, Aug 10 2017

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
  • H. J. Ryser, Combinatorial Mathematics, Carus Mathematical Monographs No. 14, John Wiley and Sons, 1963, pp. 1-8.

Crossrefs

Cf. A093563 ((6, 1) Pascal, column m=5).
Cf. A034827 (3-cycles in the triangular honeycomb bishop graph), A290775 (5-cycles), A290779 (6-cycles).

Programs

  • Mathematica
    Join[{0}, Accumulate[LinearRecurrence[{5, -10, 10, -5, 1},{1, 10, 40, 110, 245}, 40]]] (* Harvey P. Dale, Nov 30 2014 *)
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 11, 51, 161, 406}, 40] (* Harvey P. Dale, Nov 30 2014 *)
    Table[(6 n - 1) Binomial[n + 3, 4]/5, {n, 0, 20}] (* Eric W. Weisstein, Aug 10 2017 *)

Formula

a(n) = C(n+3,4) * (6*n-1)/5
G.f.: x*(1+5*x)/(1-x)^6.
a(n) = n*(n+1)*(n+2)*(n+3)*(6n-1)/120. - Derek I. Thomas (dithom02(AT)louisville.edu), Jun 30 2007

Extensions

a(1) corrected by Gael Linder (linder.gael(AT)wanadoo.fr), Oct 31 2007
a(0) prepended by Joerg Arndt, Jun 26 2013

A213500 Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.

Original entry on oeis.org

1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90
Offset: 1

Views

Author

Clark Kimberling, Jun 14 2012

Keywords

Comments

Principal diagonal: A002412.
Antidiagonal sums: A002415.
Row 1: (1,2,3,...)**(1,2,3,...) = A000292.
Row 2: (1,2,3,...)**(2,3,4,...) = A005581.
Row 3: (1,2,3,...)**(3,4,5,...) = A006503.
Row 4: (1,2,3,...)**(4,5,6,...) = A060488.
Row 5: (1,2,3,...)**(5,6,7,...) = A096941.
Row 6: (1,2,3,...)**(6,7,8,...) = A096957.
...
In general, the convolution of two infinite sequences is defined from the convolution of two n-tuples: let X(n) = (x(1),...,x(n)) and Y(n)=(y(1),...,y(n)); then X(n)**Y(n) = x(1)*y(n)+x(2)*y(n-1)+...+x(n)*y(1); this sum is the n-th term in the convolution of infinite sequences:(x(1),...,x(n),...)**(y(1),...,y(n),...), for all n>=1.
...
In the following guide to related arrays and sequences, row n of each array T(n,k) is the convolution b**c of the sequences b(h) and c(h+n-1). The principal diagonal is given by T(n,n) and the n-th antidiagonal sum by S(n). In some cases, T(n,n) or S(n) differs in offset from the listed sequence.
b(h)........ c(h)........ T(n,k) .. T(n,n) .. S(n)
h .......... h .......... A213500 . A002412 . A002415
h .......... h^2 ........ A212891 . A213436 . A024166
h^2 ........ h .......... A213503 . A117066 . A033455
h^2 ........ h^2 ........ A213505 . A213546 . A213547
h .......... h*(h+1)/2 .. A213548 . A213549 . A051836
h*(h+1)/2 .. h .......... A213550 . A002418 . A005585
h*(h+1)/2 .. h*(h+1)/2 .. A213551 . A213552 . A051923
h .......... h^3 ........ A213553 . A213554 . A101089
h^3 ........ h .......... A213555 . A213556 . A213547
h^3 ........ h^3 ........ A213558 . A213559 . A213560
h^2 ........ h*(h+1)/2 .. A213561 . A213562 . A213563
h*(h+1)/2 .. h^2 ........ A213564 . A213565 . A101094
2^(h-1) .... h .......... A213568 . A213569 . A047520
2^(h-1) .... h^2 ........ A213573 . A213574 . A213575
h .......... Fibo(h) .... A213576 . A213577 . A213578
Fibo(h) .... h .......... A213579 . A213580 . A053808
Fibo(h) .... Fibo(h) .... A067418 . A027991 . A067988
Fibo(h+1) .. h .......... A213584 . A213585 . A213586
Fibo(n+1) .. Fibo(h+1) .. A213587 . A213588 . A213589
h^2 ........ Fibo(h) .... A213590 . A213504 . A213557
Fibo(h) .... h^2 ........ A213566 . A213567 . A213570
h .......... -1+2^h ..... A213571 . A213572 . A213581
-1+2^h ..... h .......... A213582 . A213583 . A156928
-1+2^h ..... -1+2^h ..... A213747 . A213748 . A213749
h .......... 2*h-1 ...... A213750 . A007585 . A002417
2*h-1 ...... h .......... A213751 . A051662 . A006325
2*h-1 ...... 2*h-1 ...... A213752 . A100157 . A071238
2*h-1 ...... -1+2^h ..... A213753 . A213754 . A213755
-1+2^h ..... 2*h-1 ...... A213756 . A213757 . A213758
2^(n-1) .... 2*h-1 ...... A213762 . A213763 . A213764
2*h-1 ...... Fibo(h) .... A213765 . A213766 . A213767
Fibo(h) .... 2*h-1 ...... A213768 . A213769 . A213770
Fibo(h+1) .. 2*h-1 ...... A213774 . A213775 . A213776
Fibo(h) .... Fibo(h+1) .. A213777 . A001870 . A152881
h .......... 1+[h/2] .... A213778 . A213779 . A213780
1+[h/2] .... h .......... A213781 . A213782 . A005712
1+[h/2] .... [(h+1)/2] .. A213783 . A213759 . A213760
h .......... 3*h-2 ...... A213761 . A172073 . A002419
3*h-2 ...... h .......... A213771 . A213772 . A132117
3*h-2 ...... 3*h-2 ...... A213773 . A214092 . A213818
h .......... 3*h-1 ...... A213819 . A213820 . A153978
3*h-1 ...... h .......... A213821 . A033431 . A176060
3*h-1 ...... 3*h-1 ...... A213822 . A213823 . A213824
3*h-1 ...... 3*h-2 ...... A213825 . A213826 . A213827
3*h-2 ...... 3*h-1 ...... A213828 . A213829 . A213830
2*h-1 ...... 3*h-2 ...... A213831 . A213832 . A212560
3*h-2 ...... 2*h-1 ...... A213833 . A130748 . A213834
h .......... 4*h-3 ...... A213835 . A172078 . A051797
4*h-3 ...... h .......... A213836 . A213837 . A071238
4*h-3 ...... 2*h-1 ...... A213838 . A213839 . A213840
2*h-1 ...... 4*h-3 ...... A213841 . A213842 . A213843
2*h-1 ...... 4*h-1 ...... A213844 . A213845 . A213846
4*h-1 ...... 2*h-1 ...... A213847 . A213848 . A180324
[(h+1)/2] .. [(h+1)/2] .. A213849 . A049778 . A213850
h .......... C(2*h-2,h-1) A213853
...
Suppose that u = (u(n)) and v = (v(n)) are sequences having generating functions U(x) and V(x), respectively. Then the convolution u**v has generating function U(x)*V(x). Accordingly, if u and v are homogeneous linear recurrence sequences, then every row of the convolution array T satisfies the same homogeneous linear recurrence equation, which can be easily obtained from the denominator of U(x)*V(x). Also, every column of T has the same homogeneous linear recurrence as v.

Examples

			Northwest corner (the array is read by southwest falling antidiagonals):
  1,  4, 10, 20,  35,  56,  84, ...
  2,  7, 16, 30,  50,  77, 112, ...
  3, 10, 22, 40,  65,  98, 140, ...
  4, 13, 28, 50,  80, 119, 168, ...
  5, 16, 34, 60,  95, 140, 196, ...
  6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.
		

Crossrefs

Cf. A000027.

Programs

  • Mathematica
    b[n_] := n; c[n_] := n
    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}]  (* A213500 *)
  • PARI
    t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
    tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
    tabl(12) \\ Indranil Ghosh, Mar 26 2017
    
  • Python
    def t(n, k): return sum((k - i) * (n + i) for i in range(k))
    for n in range(1, 13):
        print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017

Formula

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

A061579 Reverse one number (0), then two numbers (2,1), then three (5,4,3), then four (9,8,7,6), etc.

Original entry on oeis.org

0, 2, 1, 5, 4, 3, 9, 8, 7, 6, 14, 13, 12, 11, 10, 20, 19, 18, 17, 16, 15, 27, 26, 25, 24, 23, 22, 21, 35, 34, 33, 32, 31, 30, 29, 28, 44, 43, 42, 41, 40, 39, 38, 37, 36, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66
Offset: 0

Views

Author

Henry Bottomley, May 21 2001

Keywords

Comments

A self-inverse permutation of the nonnegative numbers.
a(n) is the smallest nonnegative integer not yet in the sequence such that n + a(n) is one less than a square. - Franklin T. Adams-Watters, Apr 06 2009
From Michel Marcus, Mar 01 2021: (Start)
Array T(n,k) = (n+k)^2/2 + (n+3*k)/2 for n,k >= 0 read by descending antidiagonals.
Array T(n,k) = (n+k)^2/2 + (3*n+k)/2 for n,k >= 0 read by ascending antidiagonals. (End)

Examples

			Read as a triangle, the sequence is:
    0
    2   1
    5   4   3
    9   8   7   6
   14  13  12  11  10
  (...)
As an infinite square matrix (cf. the "table" link, 2nd paragraph) it reads:
    0    2    5    9   14   20   ...
    1    4    8   13   19   22   ...
    3    7   12   18   23   30   ...
    6   11   17   24   31   39   ...
  (...)
		

Crossrefs

Fixed points are A046092.
Row sums give A027480.
Each reversal involves the numbers from A000217 through to A000096.
Cf. A038722. Transpose of A001477.

Programs

  • Maple
    T:= (n,k)-> n*(n+3)/2-k:
    seq(seq(T(n,k), k=0..n), n=0..12);  # Alois P. Heinz, Feb 10 2023
  • Mathematica
    Module[{nn=20},Reverse/@TakeList[Range[0,(nn(nn+1))/2],Range[nn]]]// Flatten (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Jul 06 2018 *)
  • PARI
    A061579_row(n)=vector(n+=1, j, n*(n+1)\2-j)
    A061579_upto(n)=concat([A061579_row(r)|r<-[0..sqrtint(2*n)]]) \\ yields approximately n terms: actual number differs by less than +- sqrt(n). - M. F. Hasler, Nov 09 2021
    
  • Python
    from math import isqrt
    def A061579(n): return (r:=isqrt((n<<3)+1)-1>>1)*(r+2)-n # Chai Wah Wu, Feb 10 2023

Formula

a(n) = floor(sqrt(2n+1)-1/2)*floor(sqrt(2n+1)+3/2) - n = A005563(A003056(n)) - n.
Row (or antidiagonal) n = 0, 1, 2, ... contains the integers from A000217(n) to A000217(n+1)-1 in reverse order (for diagonals, "reversed" with respect to the canonical "falling" order, cf. A001477/table). - M. F. Hasler, Nov 09 2021
From Alois P. Heinz, Feb 10 2023: (Start)
T(n,k) = n*(n+3)/2 - k.
Sum_{k=0..n} k * T(n,k) = A002419(n).
Sum_{k=0..n} k^2 * T(n,k) = A119771(n).
Sum_{k=0..n} (-1)^k * T(n,k) = A226725(n). (End)

A080852 Square array of 4D pyramidal numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 5, 10, 1, 6, 15, 20, 1, 7, 20, 35, 35, 1, 8, 25, 50, 70, 56, 1, 9, 30, 65, 105, 126, 84, 1, 10, 35, 80, 140, 196, 210, 120, 1, 11, 40, 95, 175, 266, 336, 330, 165, 1, 12, 45, 110, 210, 336, 462, 540, 495, 220, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 286
Offset: 0

Views

Author

Paul Barry, Feb 21 2003

Keywords

Comments

The first row contains the tetrahedral numbers, which are really three-dimensional, but can be regarded as degenerate 4D pyramidal numbers. - N. J. A. Sloane, Aug 28 2015

Examples

			Array, n >= 0, k >= 0, begins
1 4 10 20  35  56 ...
1 5 15 35  70 126 ...
1 6 20 50 105 196 ...
1 7 25 65 140 266 ...
1 8 30 80 175 336 ...
		

Crossrefs

Cf. A057145, A080851, A180266, A055796 (antidiagonal sums).
See A257200 for another version of the array.

Programs

  • Derive
    vector(vector(poly_coeff(Taylor((1+kx)/(1-x)^5,x,11),x,n),n,0,11),k,-1,10)
    
  • Derive
    VECTOR(VECTOR(comb(k+3,3)+comb(k+3,4)n, k, 0, 11), n, 0, 11)
  • Maple
    A080852 := proc(n,k)
            binomial(k+4,4)+(n-1)*binomial(k+3,4) ;
    end proc:
    seq( seq(A080852(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Oct 01 2021
  • Mathematica
    T[n_, k_] := Binomial[k+3, 3] + Binomial[k+3, 4]n;
    Table[T[n-k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 05 2023 *)

Formula

T(n, k) = binomial(k + 4, 4) + (n-1)*binomial(k + 3, 4), corrected Oct 01 2021.
T(n, k) = T(n - 1, k) + C(k + 3, 4) = T(n - 1, k) + k(k + 1)(k + 2)(k + 3)/24.
G.f. for rows: (1 + nx)/(1 - x)^5, n >= -1.
T(n,k) = sum_{j=0..k} A080851(n,j). - R. J. Mathar, Jul 28 2016

A093563 (6,1)-Pascal triangle.

Original entry on oeis.org

1, 6, 1, 6, 7, 1, 6, 13, 8, 1, 6, 19, 21, 9, 1, 6, 25, 40, 30, 10, 1, 6, 31, 65, 70, 40, 11, 1, 6, 37, 96, 135, 110, 51, 12, 1, 6, 43, 133, 231, 245, 161, 63, 13, 1, 6, 49, 176, 364, 476, 406, 224, 76, 14, 1, 6, 55, 225, 540, 840, 882, 630, 300, 90, 15, 1, 6, 61, 280, 765, 1380
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+5*z)/(1-(1+x)*z).
The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			Triangle begins
  1;
  6,  1;
  6,  7,  1;
  6, 13,  8,  1;
  6, 19, 21,  9,  1;
  6, 25, 40, 30, 10,  1;
  ...
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.

Programs

  • Haskell
    a093563 n k = a093563_tabl !! n !! k
    a093563_row n = a093563_tabl !! n
    a093563_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Mathematica
    lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* Jean-François Alcover, Sep 16 2011, after g.f. *)
  • Python
    from math import comb, isqrt
    def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 5*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A220212 Convolution of natural numbers (A000027) with tetradecagonal numbers (A051866).

Original entry on oeis.org

0, 1, 16, 70, 200, 455, 896, 1596, 2640, 4125, 6160, 8866, 12376, 16835, 22400, 29240, 37536, 47481, 59280, 73150, 89320, 108031, 129536, 154100, 182000, 213525, 248976, 288666, 332920, 382075, 436480, 496496, 562496, 634865, 714000, 800310, 894216, 996151
Offset: 0

Views

Author

Bruno Berselli, Dec 08 2012

Keywords

Comments

Partial sums of A172073.
Apart from 0, all terms are in A135021: a(n) = A135021(A034856(n+1)) with n>0.

Crossrefs

Cf. convolution of the natural numbers (A000027) with the k-gonal numbers (* means "except 0"):
k= 2 (A000027 ): A000292;
k= 3 (A000217 ): A000332 (after the third term);
k= 4 (A000290 ): A002415 (after the first term);
k= 5 (A000326 ): A001296;
k= 6 (A000384*): A002417;
k= 7 (A000566 ): A002418;
k= 8 (A000567*): A002419;
k= 9 (A001106*): A051740;
k=10 (A001107*): A051797;
k=11 (A051682*): A051798;
k=12 (A051624*): A051799;
k=13 (A051865*): A055268.
Cf. similar sequences with formula n*(n+1)*(n+2)*(k*n-k+2)/12 listed in A264850.

Programs

  • Magma
    A051866:=func; [&+[(n-k+1)*A051866(k): k in [0..n]]: n in [0..37]];
    
  • Magma
    I:=[0,1,16,70,200]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Mathematica
    A051866[k_] := k (6 k - 5); Table[Sum[(n - k + 1) A051866[k], {k, 0, n}], {n, 0, 37}]
    CoefficientList[Series[x (1 + 11 x) / (1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2013 *)

Formula

G.f.: x*(1+11*x)/(1-x)^5.
a(n) = n*(n+1)*(n+2)*(3*n-2)/6.
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 3*(3*sqrt(3)*Pi + 27*log(3) - 17)/80.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*(6*sqrt(3)*Pi - 64*log(2) + 37)/80. (End)

A213761 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = 3*n-5+3*h, n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 6, 4, 18, 15, 7, 40, 36, 24, 10, 75, 70, 54, 33, 13, 126, 120, 100, 72, 42, 16, 196, 189, 165, 130, 90, 51, 19, 288, 280, 252, 210, 160, 108, 60, 22, 405, 396, 364, 315, 255, 190, 126, 69, 25, 550, 540, 504, 448, 378, 300
Offset: 1

Views

Author

Clark Kimberling, Jul 04 2012

Keywords

Comments

Principal diagonal: A172073.
Antidiagonal sums: A002419.
Row 1, (1,2,3,4,5,...)**(1,4,7,10,13,...): A002411.
Row 2, (1,2,3,4,5,...)**(4,7,10,13,16,...): A077414.
Row 3, (1,2,3,4,5,...)**(7,10,13,16,...): (k^3 + 7*k^2 + 6*k)/2.
Row 4, (1,2,3,4,5,...)**(10,13,16,...): (k^3 + 10*k^2 + 9*k)/2.
For a guide to related arrays, see A212500.

Examples

			Northwest corner (the array is read by falling antidiagonals):
1....6....18...40....75....126
4....15...36...70....120...189
7....24...54...100...165...252
10...33...72...130...210...315
13...42...90...160...255...378
		

Crossrefs

Cf. A212500.

Programs

  • Mathematica
    b[n_]:=n;c[n_]:=3n-2;
    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}] (* A213761 *)
    Table[t[n,n],{n,1,40}] (* A172073 *)
    s[n_]:=Sum[t[i,n+1-i],{i,1,n}]
    Table[s[n],{n,1,50}] (* A002419 *)

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*(3*n - 2 - (3*n - 5)*x) and g(x) = (1 - x)^4.

A366037 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 5 * A(x)).

Original entry on oeis.org

0, 1, 10, 160, 3110, 67155, 1548526, 37346040, 930513870, 23765376580, 618871054120, 16370119905880, 438628647940730, 11880264846822610, 324739360804852980, 8946782070689651280, 248184394985913218910, 6926162613387923126700, 194320992885495965332600, 5477763483026946993808960, 155070883903415687652796120
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 26 2023

Keywords

Comments

Reversion of g.f. for 4-dimensional figurate numbers A002419 (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^5/(1 - 5 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 5 x)/(1 + x)^5, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[5 n, n - k - 1] 5^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(5*n,n-k-1) * 5^k for n > 0.
a(n) ~ sqrt((5168 - 869*sqrt(34)) / (17*Pi)) * (22 - sqrt(34))^(5*n) / (2 * n^(3/2) * 3^(3*n + 3/2) * 5^(4*n + 1) * (11*sqrt(34) - 62)^n). - Vaclav Kotesovec, Sep 27 2023

A261721 Fourth-dimensional figurate numbers.

Original entry on oeis.org

1, 1, 5, 1, 6, 15, 1, 7, 20, 35, 1, 8, 25, 50, 70, 1, 9, 30, 65, 105, 126, 1, 10, 35, 80, 140, 196, 210, 1, 11, 40, 95, 175, 266, 336, 330, 1, 12, 45, 110, 210, 336, 462, 540, 495, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 1, 14, 55, 140, 280, 476, 714, 960, 1155, 1210, 1001, 1
Offset: 1

Views

Author

Gary W. Adamson, Aug 30 2015

Keywords

Comments

Generating polygons for the sequences are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ... .
n-th row sequence is the binomial transform of the fourth row of Pascal's triangle (1,n) followed by zeros; and the fourth partial sum of (1, n, n, n, ...).
n-th row sequence is the binomial transform of:
((n-1) * (0, 1, 3, 3, 1, 0, 0, 0) + (1, 4, 6, 4, 1, 0, 0, 0)).
Given the n-th row of the array (1, b, c, d, ...), the next row of the array is (1, b, c, d, ...) + (0, 1, 5, 15, 35, ...)

Examples

			The array as shown in A257200:
  1,  5, 15,  35,  70, 126, 210,  330, ... A000332
  1,  6, 20,  50, 105, 196, 336,  540, ... A002415
  1,  7, 25,  65, 140, 266, 462,  750, ... A001296
  1,  8, 30,  80, 175, 336, 588,  960, ... A002417
  1,  9, 35,  95, 210, 406, 714, 1170, ... A002418
  1, 10, 40, 110, 245, 476, 840, 1380, ... A002419
  ...
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of the fourth row of Pascal's triangle (1,3) followed by zeros: (1, 6, 12, 10, 3, 0, 0, 0, ...); and the fourth partial sum of (1, 3, 3, 3, 0, 0, 0).
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of: (2 * (0, 1, 3, 3, 1, 0, 0, 0, ...) + (1, 4, 6, 4, 1, 0, 0, 0, ...)); that is, the binomial transform of (1, 6, 12, 10, 3, 0, 0, 0, ...).
Row 2 of the array is (1, 5, 15, 35, 70, ...) + (0, 1, 5, 15, 35, ...), = (1, 6, 20, 50, 105, ...).
		

References

  • Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 195 (Table 80).

Crossrefs

Cf. A257200, A261720 (pyramidal numbers), A000332, A002415, A001296, A002417, A002418, A002419.
Similar to A080852 but without row n=0.
Main diagonal gives A256859.

Programs

  • Maple
    A:= (n, k)-> binomial(k+3, 3) + n*binomial(k+3, 4):
    seq(seq(A(d-k, k), k=0..d-1), d=1..13);  # Alois P. Heinz, Aug 31 2015
  • Mathematica
    row[1] = LinearRecurrence[{5, -10, 10, -5, 1}, {1, 5, 15, 35, 70}, m = 10];
    row1 = Join[{0}, row[1] // Most]; row[n_] := row[n] = row[n-1] + row1;
    Table[row[n-k+1][[k]], {n, 1, m}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 26 2016 *)
  • PARI
    A(n, k) = binomial(k+3, 3) + n*binomial(k+3, 4)
    table(n, k) = for(x=1, n, for(y=0, k-1, print1(A(x, y), ", ")); print(""))
    /* Print initial 6 rows and 8 columns as follows: */
    table(6, 8) \\ Felix Fröhlich, Jul 28 2016

Formula

G.f. for row n: (1 + (n-1)*x)/(1 - x)^5.
A(n,k) = C(k+3,3) + n * C(k+3,4) = A080852(n,k).
E.g.f. as array: exp(y)*(exp(x)*(24 + 24*(3 + x)*y + 36*(1 + x)*y^2 + 4*(1 + 3*x)*y^3 + x*y^4) - 4*(6 + 18*y + 9*y^2 + y^3))/24. - Stefano Spezia, Aug 15 2024

A264850 a(n) = n*(n + 1)*(n + 2)*(7*n - 5)/12.

Original entry on oeis.org

0, 1, 18, 80, 230, 525, 1036, 1848, 3060, 4785, 7150, 10296, 14378, 19565, 26040, 34000, 43656, 55233, 68970, 85120, 103950, 125741, 150788, 179400, 211900, 248625, 289926, 336168, 387730, 445005, 508400, 578336, 655248, 739585, 831810, 932400, 1041846
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 26 2015

Keywords

Comments

Partial sums of 16-gonal (or hexadecagonal) pyramidal numbers. Therefore, this is the case k=7 of the general formula n*(n + 1)*(n + 2)*(k*n - k + 2)/12, which is related to 2*(k+1)-gonal pyramidal numbers.

Crossrefs

Cf. A172076.
Cf. similar sequences with formula n*(n+1)*(n+2)*(k*n-k+2)/12: A000292 (k=0), A002415 (which arises from k=1), A002417 (k=2), A002419 (k=3), A051797 (k=4), A051799 (k=5), A220212 (k=6), this sequence (k=7), A264851 (k=8), A264852 (k=9).

Programs

  • Magma
    [n*(n+1)*(n+2)*(7*n-5)/12: n in [0..50]]; // Vincenzo Librandi, Nov 27 2015
    
  • Mathematica
    Table[n (n + 1) (n + 2) (7 n - 5)/12, {n, 0, 50}]
    LinearRecurrence[{5,-10,10,-5,1},{0,1,18,80,230},40] (* Harvey P. Dale, Sep 27 2018 *)
  • PARI
    a(n)=n*(n+1)*(n+2)*(7*n-5)/12 \\ Charles R Greathouse IV, Jul 26 2016

Formula

G.f.: x*(1 + 13*x)/(1 - x)^5.
a(n) = Sum_{k = 0..n} A172076(k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Nov 27 2015
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