A000127 -id:A000127 - cp's OEIS frontend

A123578 The Kruskal-Macaulay function M_2(n).

0, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13

Offset: 0

N. J. A. Sloane, Nov 12 2006

Identical to A002024, except for the initial 0.

Write n (uniquely) as n = C(n_t,t) + C(n_{t-1},t-1) + ... + C(n_v,v) where n_t > n_{t-1} > ... > n_v >= v >= 1. Then M_t(n) = C(n_t-1,t-1) + C(n_{t-1}-1,t-2) + ... + C(n_v-1,v-1).

D. E. Knuth, The Art of Computer Programming, Vol. 4, Fascicle 3, Section 7.2.1.3, Table 3.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
B. M. Abrego, S. Fernandez-Merchant, B. Llano, An Inequality for Macaulay Functions, J. Int. Seq. 14 (2011) # 11.7.4

For M_i(n), i=1, 2, 3, 4, 5 see A000127, A123578, A123579, A123580, A123731.

lowpol := proc(n,t) local x::integer ; x := floor( (n*factorial(t))^(1/t)) ; while binomial(x,t) <= n do x := x+1 ; od ; RETURN(x-1) ; end:
C := proc(n,t) local nresid,tresid,m,a ; nresid := n ; tresid := t ; a := [] ; while nresid > 0 do m := lowpol(nresid,tresid) ; a := [op(a),m] ; nresid := nresid - binomial(m,tresid) ; tresid := tresid-1 ; od ; RETURN(a) ; end:
M := proc(n,t) local a ; a := C(n,t) ; add( binomial(op(i,a)-1,t-i),i=1..nops(a)) ; end:
A123578 := proc(n) M(n,2) ; end: # R. J. Mathar, Mar 14 2007
a := proc(n) local t, s; t := 1; s := 0;
while t <= n do s := s + 1; t := t + s od; s end:
seq(a(n), n=0..84); # Peter Luschny, Oct 18 2017

lowpol[n_, t_] := Module[{x = Floor[(n*t!)^(1/t)]}, While[Binomial[x, t] <= n, x = x+1]; x-1]; c[n_, t_] := Module[{nresid = n, tresid = t, a = {}, m}, While[nresid > 0, m = lowpol[nresid, tresid]; AppendTo[a, m]; nresid = nresid - Binomial[m, tresid]; tresid = tresid-1]; a]; m[n_, t_] := With[{a = c[n, t]}, Sum[ Binomial[ a[[i]]-1, t-i], {i, 1, Length[a]}]]; a[n_] := m[n, 2]; Table[a[n], {n, 0, 84}] (* Jean-François Alcover, Dec 04 2012, translated from R. J. Mathar's Maple program *)

A123578(n)=(sqrtint(8*n)+1)\2 \\ M. F. Hasler, Apr 19 2014

from math import isqrt
def A123578(n): return isqrt(n<<3)+1>>1 # Chai Wah Wu, Oct 17 2022

A219531 a(n) = Sum_{k=0..11} C(n, k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4095, 8178, 16278, 32192, 63019, 121670, 230964, 430104, 784626, 1401292, 2449868, 4194304, 7036530, 11576916, 18696432, 29666704, 46295513, 71116846, 107636402, 160645504, 236618693, 344212906, 494889092

Offset: 0

Mokhtar Mohamed, Nov 21 2012

a(n) is the number of compositions (ordered partitions) of n+1 into twelve or fewer parts. a(n) = sum(binomial(n + 1, 2k - 1), for k = 1 .. 6). a(n) is the sum of the first twelve terms in the n-th row of Pascal's triangle.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863.

Cf. A007318.

List([0..40], n-> Sum([0..11], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019

a219531 = sum . take 12 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012

[(&+[Binomial(n,k): k in [0..11]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019

seq(sum(binomial(n,j), j=0..11), n=0..40); # G. C. Greubel, Sep 13 2019

Table[Sum[Binomial[n, k], {k, 0, 11}], {n, 0, 40}] (* T. D. Noe, Nov 23 2012 *)
LinearRecurrence[{12,-66,220,-495,792,-924,792,-495,220,-66,12,-1},{1,2,4,8,16,32,64,128,256,512,1024,2048},40] (* Harvey P. Dale, Sep 19 2019 *)

vector(40, n, sum(j=0,11, binomial(n-1,j))) \\ G. C. Greubel, Sep 13 2019

A219531_list, m = [], [1, -9, 37, -91, 148, -166, 130, -70, 25, -5, 1, 1]
for _ in range(10**2):
    A219531_list.append(m[-1])
    for i in range(11):
        m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016

[sum(binomial(n,k) for k in (0..11)) for n in (0..40)] # G. C. Greubel, Sep 13 2019

a(n) = 1 + (n^11 - 44*n^10 + 935*n^9 - 11550*n^8 + 94083*n^7 - 497112*n^6 +1870385*n^5 -3920950*n^4 +8550916*n^3 +4429656*n^2 +29400480*n)/11!. a(n) = 2*a(n - 1), for 1 <= n <= 11 with a(0) = 1, a(n) = 2*a(n - 1) - C(n - 1, 11), for n > 11. - Mohamed

G.f.: (1 - 10*x + 46*x^2 - 128*x^3 + 239*x^4 - 314*x^5 + 296*x^6 - 200*x^7 + 95*x^8 - 30*x^9 + 6*x^10)/(1-x)^12. - Mokhtar Mohamed, Nov 23 2012

A059173 Maximal number of regions into which 4-space can be divided by n hyperspheres.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 62, 114, 198, 326, 512, 772, 1124, 1588, 2186, 2942, 3882, 5034, 6428, 8096, 10072, 12392, 15094, 18218, 21806, 25902, 30552, 35804, 41708, 48316, 55682, 63862, 72914, 82898, 93876, 105912, 119072, 133424, 149038

Offset: 0

N. J. A. Sloane, Feb 15 2001

n hyperspheres divide R^k into at most C(n-1, k) + Sum_{i=0..k} C(n, i) regions.
From Raphie Frank Nov 24 2012: (Start)
Define the gross polygonal sum, GPS(n), of an n-gon as the maximal number of combined points (p), intersections (i), connections (c = edges (e) + diagonals (d)) and areas (a) of a fully connected n-gon, plus the area outside the n-gon. The gross polygonal sum (p + i + c + a + 1) is equal to this sequence and, for all n > 0, then individual components of this sum can be calculated from the first 5 entries in row (n-1) of Pascal's triangle.
For example, the gross polygonal sum of a 7-gon (the heptagon):
Let row 6 of Pascal's triangle = {1, 6, 15, 20, 15, 6, 1} = A B C D E F G.
Points = 1 + 6 = A + B = 7 [A000027(n)].
Intersections = 20 + 15 = D + E = 35 [A000332(n+2)].
Connections = 6 + 15 = B + C = 21 [A000217(n)].
Areas inside = 15 + 20 + 15 = C + D + E = 50 [A006522(n+1)].
Areas outside = 1 = A = 1 [A000012(n)].
Then, GPS(7) = 7 + 35 + 21 + 50 + 1 = 2(A + B + C + D + E) = 114 = a(7). In general, a(n) = GPS(n). (End)

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), A059174 (dim 5).
A row of A059250.
Cf. A000012, A000027, A000217, A000332, A006522.

LinearRecurrence[{5,-10,10,-5,1},2^Range[0,5],50] (* Paolo Xausa, Dec 29 2023 *)

a(0) = 1; a(n) = 2 * A000127(n), for n >= 1.
G.f.: -(x^5 + x^4 - 2*x^3 + 4*x^2 - 3*x + 1)/(x-1)^5. - Colin Barker, Oct 06 2012
E.g.f.: exp(x)*(2 + x^2 + x^4/12) - 1. - Stefano Spezia, May 19 2024

A161856 Triangle read by rows in which row n lists the coefficients of the interpolating polynomial for its divisors of n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 4, 1, 1, 0, 2, 1, 6, 1, 1, 1, 1, 1, 2, 4, 1, 1, 2, 0, 1, 10, 1, 1, 0, 0, 1, 1, 1, 12, 1, 1, 4, -2, 1, 2, 0, 8, 1, 1, 1, 1, 1, 1, 16, 1, 1, 0, 2, -4, 12, 1, 18, 1, 1, 1, -2, 7, -11, 1, 2, 2, 8, 1, 1, 8, -6, 1, 22, 1, 1, 0, 0, 1, -3, 8, -12, 1, 4, 16, 1, 1, 10, -8, 1, 2, 4, 8, 1, 1

Offset: 1

Reinhard Zumkeller, Jun 20 2009

EDP(n,x) = SUM(a(A006218(n)-1+i)*A007318(x,i-1): 1<=i<=A000005(n)) is the interpolating polynomial for the divisors of n, see also A161700;
A000005(n) = length of n-th row, i.e. same length as n-th row in A027750;
sum of n-th row, n>1: A161857(n) = SUM(a(A006218(n-1)+i): 1<=i<=A000005(n));
a(A006218(n)+1) = 1.

			1; 1,1; 1,2; 1,1,1; 1,4; 1,1,0,2; 1,6; 1,1,1,1; 1,2,4; ... .

R. Zumkeller, Table of rows 1..1000, flattened
R. Zumkeller, Enumerations of Divisors

A005408, A000124, A016813, A086514, A000125, A058331, A002522, A161701, A161702, A161703, A000127, A161704, A161706, A161707, A161708, A161710, A080856, A161711, A161712, A161713, A006261, A161715.

A057703 a(n) = n(94 + 5n + 25n^2 - 5n^3 + n^4)/120.

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 62, 119, 218, 381, 637, 1023, 1585, 2379, 3472, 4943, 6884, 9401, 12615, 16663, 21699, 27895, 35442, 44551, 55454, 68405, 83681, 101583, 122437, 146595, 174436, 206367, 242824, 284273, 331211, 384167, 443703, 510415, 584934, 667927

Offset: 0

Leonid Broukhis, Oct 24 2000

Previous name was: This sequence is the result of the question: If you have a tall building and 5 plates and you need to find the highest story from which a plate thrown does not break, what is the number of stories you can handle given n tries?

Number of compositions with at most five parts and sum at most n. - Beimar Naranjo, Mar 12 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Boardman, The Egg-Drop Numbers, Mathematics Magazine, 77 (2004), 368-372. [_Parthasarathy Nambi_, Sep 30 2009]
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A004006.

Differences form A055795 + 1 = A000127.

List([0..40], n-> n*(94+5*n+25*n^2-5*n^3+n^4)/120); # G. C. Greubel, Jun 05 2019

[n*(94+5*n+25*n^2-5*n^3+n^4)/120: n in [0..40]]; // G. C. Greubel, Jun 05 2019

seq(sum(binomial(n,k),k=1..5),n=0..38); # Zerinvary Lajos, Dec 13 2007

LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 3, 7, 15, 31}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

vector(40, n, n--; n*(94+5*n+25*n^2-5*n^3+n^4)/120) \\ G. C. Greubel, Jun 05 2019

[n*(94+5*n+25*n^2-5*n^3+n^4)/120 for n in (0..40)] # G. C. Greubel, Jun 05 2019

a(n) = n*(94 + 5*n + 25*n^2 - 5*n^3 + n^4)/120.
a(n) = Sum_{j=1..5} binomial(n, j). - Labos Elemer
G.f.: x*(1 - 3*x + 4*x^2 - 2*x^3 + x^4)/(1-x)^6. - Colin Barker, Apr 15 2012
E.g.f.: x*(120 + 60*x + 20*x^2 + 5*x^3 + x^4)*exp(x)/120. - G. C. Greubel, Jun 05 2019

More terms and formula from James Sellers, Oct 25 2000

Name changed by G. C. Greubel, Jun 06 2019

A058393 A square array based on 1^n (A000012) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 1, 2, 1, 1, 1, 2, 3, 1, 0, 1, 2, 4, 4, 1, 1, 1, 2, 4, 7, 5, 1, 0, 1, 2, 4, 8, 11, 6, 1, 1, 1, 2, 4, 8, 15, 16, 7, 1, 0, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 0, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 1, 2, 4, 8, 16, 32, 63, 99, 93, 46, 11, 1, 0

Offset: 0

Henry Bottomley, Nov 24 2000

Changing the formula by replacing T(0,2n)=T(1,n) by T(0,2n)=T(m,n) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058394, A058395, A057884, (and effectively A007318).

			Rows are (1,0,1,0,1,0,1,...), (1,1,1,1,1,1,...), (1,2,2,2,2,2,...), (1,3,4,4,4,...) etc.

Rows are A000035 (A000012 with zeros), A000012, A040000 etc. Columns are A000012, A001477, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863 etc. Diagonals include A000079, A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, etc. The triangles A008949, A054143 and A055248 also appear in the half of the array which is not powers of 2.

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(1, 1)=1, T(0, 2n)=T(1, n) and T(0, 2n+1)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2).

A123579 The Kruskal-Macaulay function M_3(n).

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 7, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 17, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 22, 23, 23, 24, 24, 24, 25, 25, 25, 25, 26, 26, 26, 26, 26, 27, 27, 27

Offset: 0

N. J. A. Sloane, Nov 12 2006

Write n (uniquely) as n = C(n_t,t) + C(n_{t-1},t-1) + ... + C(n_v,v) where n_t > n_{t-1} > ... > n_v >= v >= 1. Then M_t(n) = C(n_t-1,t-1) + C(n_{t-1}-1,t-2) + ... + C(n_v-1,v-1).
From Samuel Harkness, Sep 30 2022: (Start)
a(n) is the smallest number of balls needed on the base layer to stack n balls.
All nonrepeating terms other than a(0) occur at tetrahedral numbers + 1 (n = A000292 + 1).
The value of the nonrepeating terms other than a(0) are the Central Polygonal numbers (A000124). (End)

D. E. Knuth, The Art of Computer Programming, Vol. 4, Fascicle 3, Section 7.2.1.3, Table 3.

Samuel Harkness, Table of n, a(n) for n = 0..10000
B. M. Abrego, S. Fernandez-Merchant, and B. Llano, An Inequality for Macaulay Functions, J. Int. Seq. 14 (2011) # 11.7.4.

For M_i(n), i=1, 2, 3, 4, 5 see A000127, A123578, A123579, A123580, A123731.

Cf. A000124, A000292.

lowpol := proc(n,t) local x::integer ; x := floor( (n*factorial(t))^(1/t)) ; while binomial(x,t) <= n do x := x+1 ; od ; RETURN(x-1) ; end:
C := proc(n,t) local nresid,tresid,m,a ; nresid := n ; tresid := t ; a := [] ; while nresid > 0 do m := lowpol(nresid,tresid) ; a := [op(a),m] ; nresid := nresid - binomial(m,tresid) ; tresid := tresid-1 ; od ; RETURN(a) ; end:
M := proc(n,t) local a ; a := C(n,t) ; add( binomial(op(i,a)-1,t-i),i=1..nops(a)) ; end:
A123579 := proc(n) M(n,3) ; end:
for n from 0 to 120 do printf("%d, ",A123579(n)) ; od ; # R. J. Mathar, Mar 14 2007

c = 0; T = {0}; For[r = 1, r <= 7, r++, For[n = 1, n <= r, n++, c++; For[m = 1, m <= n, m++, AppendTo[T, c]]]]; Take[T, 75] (* Samuel Harkness, Sep 30 2022 *)

A123580 The Kruskal-Macaulay function M_4(n).

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 9, 10, 10, 10, 11, 12, 13, 13, 14, 15, 15, 16, 16, 16, 17, 18, 18, 19, 19, 19, 20, 20, 20, 20, 21, 22, 23, 23, 24, 25, 25, 26, 26, 26, 27, 28, 28, 29, 29, 29, 30, 30, 30, 30, 31, 32, 32, 33, 33, 33, 34, 34, 34, 34, 35, 35, 35, 35, 35, 36, 37, 38

Offset: 0

N. J. A. Sloane, Nov 12 2006

Write n (uniquely) as n = C(n_t,t) + C(n_{t-1},t-1) + ... + C(n_v,v) where n_t > n_{t-1} > ... > n_v >= v >= 1. Then M_t(n) = C(n_t-1,t-1) + C(n_{t-1}-1,t-2) + ... + C(n_v-1,v-1).

D. E. Knuth, The Art of Computer Programming, Vol. 4, Fascicle 3, Section 7.2.1.3, Table 3.

B. M. Abrego, S. Fernandez-Merchant, and B. Llano, An Inequality for Macaulay Functions, J. Int. Seq. 14 (2011) # 11.7.4.

For M_i(n), i=1, 2, 3, 4, 5 see A000127, A123578, A123579, A123580, A123731.

lowpol := proc(n,t) local x::integer; x := floor( (n*factorial(t))^(1/t)); while binomial(x,t) <= n do x := x+1; od; RETURN(x-1); end: C := proc(n,t) local nresid,tresid,m,a; nresid := n; tresid := t; a := []; while nresid > 0 do m := lowpol(nresid,tresid); a := [op(a),m]; nresid := nresid - binomial(m,tresid); tresid := tresid-1; od; RETURN(a); end: M := proc(n,t) local a; a := C(n,t); add( binomial(op(i,a)-1,t-i),i=1..nops(a)); end: A123580 := proc(n) M(n,4); end: for n from 0 to 120 do printf("%d, ",A123580(n)); od; # R. J. Mathar, Mar 14 2007

lowpol[n_, t_] := Module[{x = Floor[(n*t!)^(1/t)]}, While[Binomial[x, t] <= n, x = x + 1]; x - 1];
c[n_, t_] := Module[{n0 = n, t0 = t, a = {}, m}, While[n0 > 0, m = lowpol[n0, t0]; AppendTo[a, m]; n0 = n0 - Binomial[m, t0]; t0 = t0 - 1]; a];
M[n_, t_] := With[{a = c[n, t]}, Sum[Binomial[a[[i]] - 1, t - i], {i, 1, Length[a]}]];
A123580[n_] := M[n, 4];
Table[A123580[n], {n, 0, 73}] (* Jean-François Alcover, Mar 30 2023, after R. J. Mathar *)

A123731 The Kruskal-Macaulay function M_5(n).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 14, 15, 15, 15, 16, 17, 18, 19, 19, 20, 21, 22, 22, 23, 24, 24, 25, 25, 25, 26, 27, 28, 28, 29, 30, 30, 31, 31, 31, 32, 33, 33, 34, 34, 34, 35, 35, 35, 35, 36, 37, 38, 39, 39, 40, 41, 42, 42, 43, 44, 44, 45, 45, 45, 46

Offset: 0

N. J. A. Sloane, Nov 12 2006

Write n (uniquely) as n = C(n_t,t) + C(n_{t-1},t-1) + ... + C(n_v,v) where n_t > n_{t-1} > ... > n_v >= v >= 1. Then M_t(n) = C(n_t-1,t-1) + C(n_{t-1}-1,t-2) + ... + C(n_v-1,v-1).

D. E. Knuth, The Art of Computer Programming, Vol. 4, Fascicle 3, Section 7.2.1.3, Table 3.

For M_i(n), i=1, 2, 3, 4, 5 see A000127, A123578, A123579, A123580, A123731.

lowpol := proc(n,t) local x::integer; x := floor( (n*factorial(t))^(1/t)); while binomial(x,t) <= n do x := x+1; od; RETURN(x-1); end:
C := proc(n,t) local nresid,tresid,m,a; nresid := n; tresid := t; a := []; while nresid > 0 do m := lowpol(nresid,tresid); a := [op(a),m]; nresid := nresid - binomial(m,tresid); tresid := tresid-1; od; RETURN(a); end:
M := proc(n,t) local a; a := C(n,t); add( binomial(op(i,a)-1,t-i),i=1..nops(a)); end:
A123731 := proc(n) M(n,5); end:
for n from 0 to 120 do printf("%d, ",A123731(n)); od; # R. J. Mathar, Mar 14 2007

lowpol[n_, t_] := Module[{x = Floor[(n*t!)^(1/t)]}, While[Binomial[x, t] <= n, x = x + 1]; x - 1];
c[n_, t_] := Module[{n0 = n, t0 = t, a = {}, m}, While[n0 > 0, m = lowpol[n0, t0]; AppendTo[a, m]; n0 = n0 - Binomial[m, t0]; t0 = t0 - 1]; a];
M[n_, t_] := With[{a = c[n, t]}, Sum[Binomial[a[[i]] - 1, t - i], {i, 1, Length[a]}]];
A123731[n_] := M[n, 5];
Table[A123731[n], {n, 0, 72}] (* Jean-François Alcover, Mar 30 2023, after R. J. Mathar *)

More terms from R. J. Mathar, Mar 14 2007

A101338 Antidiagonal sums in A101321.

Original entry on oeis.org

1, 2, 4, 9, 20, 41, 77, 134, 219, 340, 506, 727, 1014, 1379, 1835, 2396, 3077, 3894, 4864, 6005, 7336, 8877, 10649, 12674, 14975, 17576, 20502, 23779, 27434, 31495, 35991, 40952, 46409, 52394, 58940, 66081, 73852, 82289, 91429, 101310, 111971

Offset: 0

Eugene McDonnell (eemcd(AT)mac.com), Dec 24 2004

Equals binomial transform of [1, 1, 1, 2, 1, 0, 0, 0, ...]. Example: a(5) = 20 = [1, 1, 1, 2, 1] dot [1, 4, 6, 4, 1] = (1 + 4 + 6 + 8 + 1). - Gary W. Adamson, Aug 25 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5, 1).

Cf. A000127, A000292.

CoefficientList[Series[(1-3*x+4*x^2-x^3)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, Mar 24 2012 *)
LinearRecurrence[{5,-10,10,-5,1},{1,2,4,9,20},50] (* Harvey P. Dale, May 21 2013 *)

a(n) = n^4/24 + n^3/12 - n^2/24 + 11*n/12 + 1.
G.f.: (1-3*x+4*x^2-x^3)/(1-x)^5. - Colin Barker, Mar 22 2012
a(0)=1, a(1)=2, a(2)=4, a(3)=9, a(4)=20, a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, May 21 2013
a(n) = A000127(n+1) + A000292(n-2). - Bruce J. Nicholson, Jan 06 2019

cp's OEIS Frontend

A123578 The Kruskal-Macaulay function M_2(n).

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A219531 a(n) = Sum_{k=0..11} C(n, k).

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A059173 Maximal number of regions into which 4-space can be divided by n hyperspheres.

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A161856 Triangle read by rows in which row n lists the coefficients of the interpolating polynomial for its divisors of n.

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A057703 a(n) = n*(94 + 5*n + 25*n^2 - 5*n^3 + n^4)/120.

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A058393 A square array based on 1^n (A000012) with each term being the sum of 2 consecutive terms in the previous row.

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A123579 The Kruskal-Macaulay function M_3(n).

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A057703 a(n) = n(94 + 5n + 25n^2 - 5n^3 + n^4)/120.