A005900 -id:A005900 - cp's OEIS frontend

A099196 a(n) = n(2n^8 + 84n^6 + 798n^4 + 1636*n^2 + 315)/2835.

0, 1, 18, 163, 996, 4645, 17718, 57799, 166344, 432073, 1030490, 2286955, 4772780, 9446125, 17852030, 32398735, 56730512, 96220561, 158611106, 254831667, 400030580, 614859189, 927052742, 1373356887, 2001853784, 2874747225, 4071671786, 5693596923, 7867403068, 10751213181

Offset: 0

Jonathan Vos Post, Nov 16 2004

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099195 (m=8), A099197 (m=10).

Cf. A000332.

concat(0, Vec(x*(1+x)^8/(1-x)^10 + O(x^40))) \\ Michel Marcus, Dec 14 2015

a(n) = n*(2*n^8 + 84*n^6 + 798*n^4 + 1636*n^2 + 315)/2835.
G.f.: x*(1+x)^8/(1-x)^10. [Colin Barker, May 01 2012]
a(n) = 18*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

More terms from Michel Marcus, Dec 14 2015

A099195 a(n) = (n^2)( n^6 + 28n^4 + 154*n^2 + 132 )/315.

Original entry on oeis.org

0, 1, 16, 129, 704, 2945, 10128, 29953, 78592, 187137, 411280, 845185, 1640640, 3032705, 5373200, 9173505, 15158272, 24331777, 38058768, 58161793, 87037120, 127791489, 184402064, 261902081, 366594816, 506298625, 690625936, 931299201, 1242506944, 1641303169, 2148053520

Offset: 0

Jonathan Vos Post, Nov 16 2004

H. S. M. Coxeter, Regular Polytopes, New York: Dover, 1973.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099196 (m=9), A099197 (m=10).

Cf. A000332.

LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{0,1,16,129,704,2945,10128,29953,78592},40] (* Harvey P. Dale, Jan 23 2019 *)

concat(0, Vec(x*(1+x)^7/(1-x)^9 + O(x^40))) \\ Michel Marcus, Dec 14 2015

a(n) = (n^2)*( n^6 + 28*n^4 + 154*n^2 + 132 )/315.
G.f.: x*(1+x)^7/(1-x)^9. [R. J. Mathar, Jul 18 2009]
a(n) = 16*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 19, 8, 1, 6, 25, 44, 33, 10, 1, 7, 36, 85, 96, 51, 12, 1, 8, 49, 146, 225, 180, 73, 14, 1, 9, 64, 231, 456, 501, 304, 99, 16, 1, 10, 81, 344, 833, 1182, 985, 476, 129, 18, 1, 11, 100, 489, 1408, 2471, 2668, 1765, 704, 163, 20, 1, 12

Offset: 0

Gary W. Adamson, Mar 19 2005

The n-th column of the triangle is the binomial transform of the n-th row of A081277, followed by zeros. Example: column 3, (1, 6, 19, 44, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). A104698 = reversal by rows of A142978. - Gary W. Adamson, Jul 17 2008
This sequence is jointly generated with A210222 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) + 1 and v(n,x) = 2x*u(n-1,x) + v(n-1,x) + 1. See the Mathematica section at A210222. - Clark Kimberling, Mar 19 2012
This Riordan triangle T appears in a formula for A001100(n, 0) = A002464(n), for n >= 1. - Wolfdieter Lang, May 13 2025

			The Riordan triangle T begins:
  n\k  0   1   2    3    4    5    6   7   8  9 10 ...
  ----------------------------------------------------
  0:   1
  1:   2   1
  2:   3   4   1
  3:   4   9   6    1
  4:   5  16  19    8    1
  5:   6  25  44   33   10    1
  6:   7  36  85   96   51   12    1
  7:   8  49 146  225  180   73   14   1
  8:   9  64 231  456  501  304   99  16   1
  9:  10  81 344  833 1182  985  476 129  18  1
  10: 11 100 489 1408 2471 2668 1765 704 163 20  1
  ... reformatted and extended by _Wolfdieter Lang_, May 13 2025
From _Wolfdieter Lang_, May 13 2025: (Start)
Zumkeller recurrence (adapted for offset [0,0]): 19 = T(4, 2) = T(2, 1) + T(3, 1) + T(3,3) = 4 + 9 + 6 = 19.
A-sequence recurrence: 19 = T(4, 2) = 1*T(3. 1) + 2*T(3. 2) - 2*T(3, 3) = 9 + 12 - 2 = 19.
Z-sequence recurrence: 5 = T(4, 0) = 2*T(3, 0) - 1*T(3, 1) + 2*T(3, 2) - 6*T(3, 3) = 8 - 9 + 12 + 6 = 5.
Boas-Buck recurrence: 19 = T(4, 2) = (1/2)*((2 + 0)*T(2, 2) + (2 + 2*2)*T(3, 2)) = (1/2)*(2 + 36) = 19. (End)

Reinhard Zumkeller, First 100 rows of triangle, flattened

Diagonal sums are A008937(n+1).
Cf. A048739 (row sums), A008288, A005900 (column 3), A014820 (column 4)
Cf. A081277, A142978 by antidiagonals, A119328, A110271 (matrix inverse).
Cf. A001100, A002464, A010673, A040000, A112478, A133872, A383857.

a104698 n k = a104698_tabl !! (n-1) !! (k-1)
a104698_row n = a104698_tabl !! (n-1)
a104698_tabl = [1] : [2,1] : f [1] [2,1] where
   f us vs = ws : f vs ws where
     ws = zipWith (+) ([0] ++ us ++ [0]) $
          zipWith (+) ([1] ++ vs) (vs ++ [0])
-- Reinhard Zumkeller, Jul 17 2015

A104698 := proc(n, k) add(binomial(k, j)*binomial(n-j+1, n-k-j), j=0..n-k) ; end proc:
seq(seq(A104698(n, k), k=0..n), n=0..15); # R. J. Mathar, Sep 04 2011
T := (n, k) -> binomial(n + 1, k + 1)*hypergeom([-k, k - n], [-n - 1], -1):
for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
T := proc(n, k) option remember; if k = 0 then n + 1 elif k = n then 1 else T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) fi end: # Peter Luschny, May 13 2025

u[1, ] = 1; v[1, ] = 1;
u[n_, x_] := u[n, x] = x u[n-1, x] + v[n-1, x] + 1;
v[n_, x_] := v[n, x] = 2 x u[n-1, x] + v[n-1, x] + 1;
Table[CoefficientList[u[n, x], x], {n, 1, 11}] // Flatten (* Jean-François Alcover, Mar 10 2019, after Clark Kimberling *)

T(n,k)=sum(j=0,n-k,binomial(k,j)*binomial(n-j+1,k+1)) \\ Charles R Greathouse IV, Jan 16 2012

The triangle is extracted from the product A * B; A = [1; 1, 1; 1, 1, 1; ...], B = [1; 1, 1; 1, 3, 1; 1, 5, 5, 1; ...] both infinite lower triangular matrices (rest of the terms are zeros). The triangle of matrix B by rows = A008288, Delannoy numbers.
From Paul Barry, Jul 18 2005: (Start)
Riordan array (1/(1-x)^2, x(1+x)/(1-x)) = (1/(1-x), x)*(1/(1-x), x(1+x)/(1-x)).
T(n, k) = Sum_{j=0..n} Sum_{i=0..j-k} C(j-k, i)*C(k, i)*2^i.
T(n, k) = Sum_{j=0..k} Sum_{i=0..n-k-j} (n-k-j-i+1)*C(k, j)*C(k+i-1, i). (End)
T(n, k) = binomial(n+1, k+1)*2F1([-k, k-n], [-n-1], -1) where 2F1 is a Gaussian hypergeometric function. - R. J. Mathar, Sep 04 2011
T(n, k) = T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) for 1 < k < n; T(n, 0) = n + 1; T(n, n) = 1. - Reinhard Zumkeller, Jul 17 2015
From Wolfdieter Lang, May 13 2025: (Start)
The Riordan triangle T = (1/(1 - x)^2, x*(1 + x)/(1 - x)) has the o.g.f. G(x, y) = 1/((1 - x)*(1 - x - y*x*(1+x))) for the row polynomials R(n, y) = Sum_{k=0..n} T(n, k)*y^k.
The o.g.f. for column k is G(k, x) = (1/(1 - x)^2)*(x*(1 + x)/(1 - x))^k, for k >= 0.
The o.g.f. for the diagonal m is D(m, x) = N(m, x)/(1 - x)^(m+1), with the numerator polynomial N(m, x) = Sum_{k=0..floor(m/2)} A034867(m, k)*x^(2*k) for m >= 0.
The row sums with o.g.f. R(x) = 1/((1 -x)*(1 - 2*x -x^2) give A048739.
The alternating row sums with o.g.f. 1/((1 - x)(1 + x^2)) give A133872.
The A-sequence for this Riordan triangle has o.g.f. A(x) = 1 + x + sqrt(1 + 6*x + x^2))/2 giving A112478(n). Hence T(n, k) = Sum_{j=0..n-k} A112478(j)*T(n-1, k-1+j), for n >= 1, k >= 1, T(n, k) = 0 for n < k, and T(0, 0) = 1.
The Z-sequence has o.g.f. (5 + x - sqrt(1 + 6*x + x^2))/2 = 3 + x - A(x) giving Z(n) = {2, -1, -A112478(n >= 2)}. Hence T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1. For A- and Z-sequences of Riordan triangles see a W. Lang link at A006232 with references.
The Boas-Buck sequences alpha and beta for the Riordan triangle T (see A046521 for the Aug 10 2017 comment and reference) are alpha(n) = A040000(n+1) = repeat{2} and beta(n) = A010673(n+1) = repeat{2,0}. Hence the recurrence for column T(n, k){n>=k}, with input T(k, k) = 1, for k >= 0, is T(n, k) = (1/(n-k)) * Sum{j=k..n-1} (2 + k*(1 + (-1)^(n-1-j))) *T(j,k), for n >= k+1. (End)

A099197 a(n) = (n^2)(2n^8+120n^6+1806n^4+7180*n^2+5067)/14175.

Original entry on oeis.org

0, 1, 20, 201, 1360, 7001, 29364, 104881, 329024, 927441, 2390004, 5707449, 12767184, 26986089, 54284244, 104535009, 193664256, 346615329, 601446996, 1014889769, 1669752016, 2684641785, 4226553716, 6526963345, 9902174016, 14778775025, 21725194036, 31490462745

Offset: 0

Jonathan Vos Post, Nov 16 2004

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099195 (m=8), A099196 (m=9).

Cf. A000332.

A099197[n_] := n^2*(2*n^2*(n^6 + 60*n^4 + 903*n^2 + 3590) + 5067)/14175;
Array[A099197, 30, 0] (* Paolo Xausa, Jan 29 2025 *)

a(n)=n^2*(2*n^8+120*n^6+1806*n^4+7180*n^2+5067)/14175 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = (n^2)*(2*n^8+120*n^6+1806*n^4+7180*n^2+5067)/14175.

A068236 First differences of (n+1)^5-n^5.

Original entry on oeis.org

30, 180, 570, 1320, 2550, 4380, 6930, 10320, 14670, 20100, 26730, 34680, 44070, 55020, 67650, 82080, 98430, 116820, 137370, 160200, 185430, 213180, 243570, 276720, 312750, 351780, 393930, 439320, 488070, 540300, 596130, 655680, 719070, 786420, 857850, 933480

Offset: 0

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Mar 25 2002

For n>=0, a(n) is equal to the number of functions f:{1,2,3,4,5}->{1,2,...,n+2} such that Im(f) contains 2 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Feb 24 2007

Colin Barker, Table of n, a(n) for n = 0..1000
O. Bagdasar, On some functions involving the lcm and gcd of integer tuples, Scientific Publications of the State University of Novi Pazar, Appl. Maths. Inform. and Mech., Vol. 6, 2 (2014), 91--100.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A022521 ((n+1)^5-n^5), A000584 (5th powers), A005900 (octahedral numbers).

Table[20*n^3 + 10*n, {n, 1, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *)
Differences[#[[2]]-#[[1]]&/@Partition[Range[0,40]^5,2,1]] (* or *) LinearRecurrence[{4,-6,4,-1},{30,180,570,1320},40] (* Harvey P. Dale, Jun 05 2019 *)

Vec(30*(x+1)^2 / (x-1)^4 + O(x^100)) \\ Colin Barker, Dec 13 2014

a(n) = (n+2)^5-2*(n+1)^5+n^5.
a(n) = 30*A005900(n+1). - R. J. Mathar, Sep 02 2008
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Colin Barker, Dec 13 2014
G.f.: 30*(x+1)^2 / (x-1)^4. - Colin Barker, Dec 13 2014

A061317 Split positive integers into extending even groups and sum: 1+2, 3+4+5+6, 7+8+9+10+11+12, 13+14+15+16+17+18+19+20, ...

Original entry on oeis.org

0, 3, 18, 57, 132, 255, 438, 693, 1032, 1467, 2010, 2673, 3468, 4407, 5502, 6765, 8208, 9843, 11682, 13737, 16020, 18543, 21318, 24357, 27672, 31275, 35178, 39393, 43932, 48807, 54030, 59613, 65568, 71907, 78642, 85785, 93348, 101343, 109782

Offset: 0

Henry Bottomley, Feb 13 2002

5*a(n+1) is the sum of the products of the 10 distinct combinations of three consecutive numbers starting with n (using 1,2,3 the 10 combinations are 111 112 113 122 123 133 222 223 233 333; 1*1*1 + 1*1*2 + 1*1*3 + 1*2*2 + 1*2*3 + 1*3*3 + 2*2*2 + 2*2*3 + 2*3*3 + 3*3*3 = 90 = 5*a(2)). - J. M. Bergot, Mar 28 2014 [expanded by Jon E. Schoenfield, Feb 22 2015]

			1+2 = 3; 3+4+5+6 = 18; 7+8+9+10+11+12 = 57; 13+14+15+16+17+18+19+20 = 132.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A005898, A006003.

Cf. A000217, A002378, A110450.

A061317:=n->2*n^3+n; seq(A061317(n), n=0..100); # Wesley Ivan Hurt, Mar 20 2014

Table[2n^3+n,{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Dec 04 2010 *)
LinearRecurrence[{4,-6,4,-1},{0,3,18,57},40] (* Harvey P. Dale, Aug 23 2015 *)
With[{nn=40},Total/@TakeList[Range[nn+nn^2],2Range[0,nn]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Mar 10 2018 *)

a(n) = { 2*n^3 + n } \\ Harry J. Smith, Jul 21 2009

a(n) = 2*n^3 + n.
a(n) = A000217(A002378(n)) - A000217(A002378(n-1)).
a(n) = 3 * A005900(n).
a(n) = A001477(n) * A058331(n).
a(n) = A000578(n) + A034262(n).
G.f.: 3*x*(1+x)^2/(x-1)^4.
a(n) = A110450(n) - A110450(n-1). - J.S. Seneschal, Jul 01 2025

A069764 Frobenius number of the numerical semigroup generated by consecutive octahedral numbers.

Original entry on oeis.org

89, 773, 3611, 12179, 33349, 78889, 167383, 326471, 595409, 1027949, 1695539, 2690843, 4131581, 6164689, 8970799, 12769039, 17822153, 24441941, 32995019, 43908899, 57678389, 74872313, 96140551, 122221399, 153949249, 192262589, 238212323, 292970411, 357838829

Offset: 2

Victoria A Sapko (vsapko(AT)canes.gsw.edu), Apr 18 2002

The Frobenius number of a numerical semigroup generated by relatively prime integers a_1,...,a_n is the largest positive integer that is not a nonnegative linear combination of a_1,...,a_n. Since consecutive octahedral numbers are relatively prime, they generate a numerical semigroup with a Frobenius number. The Frobenius number of a 2-generated semigroup has the formula ab-a-b.

			a(2)=89 because 89 is not a nonnegative linear combination of 6 and 19 (the second and third octahedral numbers), but all integers greater than 89 are.

Harvey P. Dale, Table of n, a(n) for n = 2..1000
R. Fröberg, C. Gottlieb and R. Häggkvist, On numerical semigroups, Semigroup Forum, 35 (1987), 63-83 (for definition of Frobenius number).
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A005900, A037165, A059769, A069755-A069763.

FrobeniusNumber/@Partition[Rest[Table[(n(2n^2+1))/3,{n,30}]],2,1] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{89,773,3611,12179,33349,78889,167383},30] (* Harvey P. Dale, Nov 19 2015 *)

a(n) = ((1/3)*n*(2*n^2+1)-1)*((1/3)*(n+1)*(2*(n+1)^2+1)-1)-1.
G.f.: x^2*(89+150*x+69*x^2+20*x^3-13*x^4+6*x^5-x^6)/(1-x)^7. - Colin Barker, Feb 12 2012
a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+ 21*a(n-5)- 7*a(n-6)+a(n-7). - Harvey P. Dale, Nov 19 2015

More terms from Carl Najafi, Sep 10 2011

A100185 Structured meta-anti-prism numbers, the n-th number from a structured n-gonal anti-prism number sequence.

Original entry on oeis.org

1, 4, 19, 68, 185, 416, 819, 1464, 2433, 3820, 5731, 8284, 11609, 15848, 21155, 27696, 35649, 45204, 56563, 69940, 85561, 103664, 124499, 148328, 175425, 206076, 240579, 279244, 322393, 370360, 423491, 482144

Offset: 1

James A. Record (james.record(AT)gmail.com)

			There are no 1- or 2-gonal anti-prisms, so 1 and (2n) are used as the first and second terms since all the sequences begin as such.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A005900, A000447, A096000, A100178, A100157, A100185 - structured anti-prisms; A006484 for other structured meta numbers; and A100145 for more on structured numbers.

[(1/6)*(3*n^4-8*n^3+9*n^2+2*n): n in [1..40]]; // Vincenzo Librandi, Aug 03 2011

a(n) = (1/6)*(3*n^4 - 8*n^3 + 9*n^2 + 2*n).

G.f.: x*(1 - x + 9*x^2 + 3*x^3)/(1-x)^5. [Colin Barker, Jun 08 2012]

A008867 Triangle of truncated triangular numbers: k-th term in n-th row is number of dots in hexagon of sides k, n-k, k, n-k, k, n-k.

Original entry on oeis.org

1, 3, 3, 6, 7, 6, 10, 12, 12, 10, 15, 18, 19, 18, 15, 21, 25, 27, 27, 25, 21, 28, 33, 36, 37, 36, 33, 28, 36, 42, 46, 48, 48, 46, 42, 36, 45, 52, 57, 60, 61, 60, 57, 52, 45, 55, 63, 69, 73, 75, 75, 73, 69, 63, 55, 66, 75, 82, 87, 90, 91, 90, 87, 82, 75, 66, 78, 88, 96, 102, 106, 108, 108, 106, 102, 96, 88, 78

Offset: 2

N. J. A. Sloane, J. H. Conway and R. K. Guy

Closely related to A109439. The current sequence is made of truncated triangular numbers, the latter gives the full description. Both can help to build a cube with layers perpendicular to the great diagonal. E.g.: 15,18,19,18,15 in A008867 is a truncation of the lesser triangular numbers of 1,3,6,10,15,18,19,18,15,10,6,3,1 in A109439. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 02 2005

The sequence is a triangle read by rows where the n-th row is obtained by multiplying by (1/3)*(n+1)*(2*(n+1)^2+1) the first row of the limit as k approaches infinity of P(n)^k where P(n) is the stochastic matrix associated with a variant of the Ehrenfest model using n balls. The elements of the stochastic matrix P(n) we have considered are given by P(n)[i,j] = n+1-(max(i,j)-min(i,j)), where each row must be normalized using the L1 norm and where i,j belong to the set {0,1,2,...,n}. They are defined as the probabilities of arriving in a state j given the previous state i. In particular the sum of every row of a stochastic matrix must be 1, and so the sum of the terms of the n-th row of this triangle is (1/3)*(n+1)*(2*(n+1)^2+1) (since the limit of a stochastic matrix is again a stochastic matrix). Furthermore, by the properties of Markov chains, we can interpret P(n)^k as the k-step transition matrix of this variant of the Ehrenfest model using n balls. It is important to note that the rows of the limit of the stochastic matrix are identical and since we know the first we know all the others. - Luca Onnis, Oct 29 2023

			Triangle begins:
n = 0:  1;
n = 1:  3,  3;
n = 2:  6,  7,  6;
n = 3: 10, 12, 12, 10;
n = 4: 15, 18, 19, 18, 15;
n = 5: 21, 25, 27, 27, 25, 21;
n = 6: 28, 33, 36, 37, 36, 33, 28;

Paul and Tatjana Ehrenfest, Über zwei bekannte Einwände gegen das Boltzmannsche H-Theorem, Physikalische Zeitschrift, vol. 8 (1907), pp. 311-314.

Hugo Pfoertner, Table of n, a(n) for n = 2..1226, rows 1..50, flattened.
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
Luca Onnis, Animation of this variant of the Ehrenfest model using n = 15 balls.
Wikipedia, Ehrenfest model.

Row sums are A005900(n-1).

Cf. A109439.

T:= (n, k)-> n*(n-3)/2 - k^2+k*n+1:
seq(seq(T(n,k), k=1..n-1), n=2..14);

T[n_,k_] := n*(n-3)/2 - k^2 + k*n + 1; Table[T[n,k], {n,3,20}, {k,n,2,-1}] // Flatten (* Amiram Eldar, Dec 12 2018 *)

T(n,k) = n*(n-3)/2 - k^2 + k*n + 1.

A016687 Decimal expansion of log(64) = 6*log(2).

Original entry on oeis.org

4, 1, 5, 8, 8, 8, 3, 0, 8, 3, 3, 5, 9, 6, 7, 1, 8, 5, 6, 5, 0, 3, 3, 9, 2, 7, 2, 8, 7, 4, 9, 0, 5, 9, 4, 0, 8, 4, 5, 3, 0, 0, 0, 8, 0, 6, 1, 6, 1, 5, 3, 1, 5, 2, 4, 7, 2, 4, 0, 8, 0, 0, 5, 6, 9, 6, 0, 3, 6, 1, 7, 3, 1, 8, 1, 8, 1, 6, 8, 2, 9, 3, 6, 3, 5, 1, 7, 9, 9, 6, 1, 9, 7, 8, 5, 1, 2, 1, 2

Offset: 1

N. J. A. Sloane

			4.158883083359671856503392728749059408453000806161531524724080056960361...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 2.

Harry J. Smith, Table of n, a(n) for n = 1..20000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Index entries for transcendental numbers

Cf. A002162, A005900, A016492 (continued fraction), A016627, A016631.

RealDigits[Log[64],10,120][[1]] (* Harvey P. Dale, May 06 2022 *)

default(realprecision, 20080); x=log(64); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b016687.txt", n, " ", d)); \\ Harry J. Smith, May 22 2009

Equals 2*A016631 = 3*A016627 = 6*A002162. - Alois P. Heinz, Aug 07 2023
From Peter Bala, Mar 05 2024: (Start)
log(64) = 4 + Sum_{n >= 1} (-1)^(n+1)/(p(n)*p(n+1)), where p(n) = n*(2*n^2 + 1)/3 = A005900.
Continued fraction: log(64) = 4 + 1/(6 + (1*2)/(6 + (2*3)/(6 + (3*4)/(6 + (4*5)/(6 + ... ))))). See A142983. Cf. A016627. (End)

cp's OEIS Frontend

A099196 a(n) = n*(2*n^8 + 84*n^6 + 798*n^4 + 1636*n^2 + 315)/2835.

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A099195 a(n) = (n^2)*( n^6 + 28*n^4 + 154*n^2 + 132 )/315.

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A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).

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A099197 a(n) = (n^2)*(2*n^8+120*n^6+1806*n^4+7180*n^2+5067)/14175.

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A068236 First differences of (n+1)^5-n^5.

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A061317 Split positive integers into extending even groups and sum: 1+2, 3+4+5+6, 7+8+9+10+11+12, 13+14+15+16+17+18+19+20, ...

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A069764 Frobenius number of the numerical semigroup generated by consecutive octahedral numbers.

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A099196 a(n) = n(2n^8 + 84n^6 + 798n^4 + 1636*n^2 + 315)/2835.

A099195 a(n) = (n^2)( n^6 + 28n^4 + 154*n^2 + 132 )/315.

A099197 a(n) = (n^2)(2n^8+120n^6+1806n^4+7180*n^2+5067)/14175.