A006261 -id:A006261 - cp's OEIS frontend

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

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.

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).

A374378 Iterated rascal triangle R2: T(n,k) = Sum_{m=0..2} binomial(n-k,m)*binomial(k,m).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 19, 15, 6, 1, 1, 7, 21, 31, 31, 21, 7, 1, 1, 8, 28, 46, 53, 46, 28, 8, 1, 1, 9, 36, 64, 81, 81, 64, 36, 9, 1, 1, 10, 45, 85, 115, 126, 115, 85, 45, 10, 1, 1, 11, 55, 109, 155, 181, 181, 155, 109, 55, 11, 1

Offset: 0

Kolosov Petro, Jul 06 2024

Triangle T(n,k) is the second triangle R2 among the rascal-family triangles; A374452 is triangle R3; A077028 is triangle R1.
Triangle T(n,k) equals Pascal's triangle A007318 through row 2i+1, i=2 (i.e., row 5).
Triangle T(n,k) equals Pascal's triangle A007318 through column i, i=2 (i.e., column 2).

			Triangle begins:
--------------------------------------------------
k=     0   1   2   3    4    5    6   7   8   9 10
--------------------------------------------------
n=0:   1
n=1:   1   1
n=2:   1   2   1
n=3:   1   3   3   1
n=4:   1   4   6   4    1
n=5:   1   5  10  10    5    1
n=6:   1   6  15  19   15    6    1
n=7:   1   7  21  31   31   21    7   1
n=8:   1   8  28  46   53   46   28   8   1
n=9:   1   9  36  64   81   81   64  36   9   1
n=10:  1  10  45  85  115  126  115  85  45  10  1

Kolosov Petro, Table of n, a(n) for n = 0..1325
Amelia Gibbs and Brian K. Miceli, Two Combinatorial Interpretations of Rascal Numbers, arXiv:2405.11045 [math.CO], 2024.
Jena Gregory, Brandt Kronholm, and Jacob White, Iterated rascal triangles, Aequationes mathematicae, 2023.
Jena Gregory, Iterated rascal triangles, Theses and Dissertations. 1050., The University of Texas Rio Grande Valley, 2022.
Philip K. Hotchkiss, Student Inquiry and the Rascal Triangle, arXiv:1907.07749 [math.HO], 2019.
Philip K. Hotchkiss, Generalized Rascal Triangles, Journal of Integer Sequences, Vol. 23, 2020.
Petro Kolosov, Identities in Iterated Rascal Triangles, 2024.

Cf. A077028, A007318, A000027, A000217, A005448, A056108, A212656, A000292, A095667, A095666, A006261.

t[n_, k_]:=Sum[Binomial[n - k, m]*Binomial[k, m], {m, 0, 2}]; Column[Table[t[n, k], {n, 0, 12}, {k, 0, n}], Center]

T(n,k) = 1 + k*(n-k) + (1/4)*(k-1)*k*(n-k-1)*(n-k).
Row sums give A006261(n).
Diagonal T(n+1, n) gives A000027(n).
Diagonal T(n+2, n) gives A000217(n).
Diagonal T(n+3, n) gives A005448(n).
Diagonal T(n+4, n) gives A056108(n).
Diagonal T(n+5, n) gives A212656(n).
Column k=3 difference binomial(n+6, 3) - T(n+6, 3) gives C(n+3,3)=A007318(n+3,3).
Column k=4 difference binomial(n+7, 4) - T(n+7, 4) gives fifth column of (1,4)-Pascal triangle A095667.
G.f.: (1 + 3*x^4*y^2 - (2*x + 3*x^3*y)*(1 + y) + x^2*(1 + 5*y + y^2))/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Jul 09 2024

A219615 a(n) = Sum_{k=0..12} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16369, 32647, 64839, 127858, 249528, 480492, 910596, 1695222, 3096514, 5546382, 9740686, 16777216, 28354132, 47050564, 76717268, 123012781, 194129627, 301766029, 462411533, 699030226, 1043243132

Offset: 0

Mokhtar Mohamed, Nov 23 2012

a(n) is the number of compositions (ordered partitions) of n+1 into thirteen or fewer parts.

a(n) is the sum of the first thirteen terms in the n-th row of Pascal's triangle.

			a(13)= 8191 because there are (2^13) -1 compositions of 14 into thirteen or fewer parts. When 1<= n <= 12, for n=5, a(5) = 2*a(4) = 2*16 = 32. For n=12, a(12) = 2*a(11)= 2*2048 = 4096. When n>12, for n=13, a(13) = 2*a(12) - binomial(12,12) = 2*4096 - 1 = 8191. For n = 15, a(15) = 2*a(14) - binomial(14,12) = 2*16369 - 91 = 32738 - 91 = 32647.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863, A219531.

Table[Sum[Binomial[n, k], {k, 0, 12}], {n, 0, 40}] (* T. D. Noe, Nov 27 2012 *)
LinearRecurrence[{13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1},{1,2,4,8,16,32,64,128,256,512,1024,2048,4096},40] (* Harvey P. Dale, Nov 29 2012 *)

a(n)=sum(k=1,12,binomial(n,k)) \\ Charles R Greathouse IV, Nov 27 2012

a(n) = (n^12 - 54n^11 + 1397n^10 - 21450n^9 + 218823n^8 - 1508562n^7 + 7374191n^6 - 23551110n^5 + 58206676n^4 - 48306984n^3 + 173699712n^2 + 312888960n)/479001600. - Charles R Greathouse IV, Nov 27 2012

a(0)=1, a(1)=2, a(2)=4, a(3)=8, a(4)=16, a(5)=32, a(6)=64, a(7)=128, a(8)=256, a(9)=512, a(10)=1024, a(11)=2048, a(12)=4096, a(n)= 13*a(n-1)- 78*a(n-2)+286*a(n-3)-715*a(n-4)+1287*a(n-5)-1716*a(n-6)+ 1716*a(n-7)- 1287*a(n-8)+715*a(n-9)-286*a(n-10)+78*a(n-11)-13*a(n-12)+a(n-13). - Harvey P. Dale, Nov 29 2012

Sequence corrected and extended by T. D. Noe, Nov 26 2012

Definition corrected by Harvey P. Dale, Nov 29 2012

A219676 a(n) = Sum_{k=0..13} binomial(n, k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16383, 32752, 65399, 130238, 258096, 507624, 988116, 1898712, 3593934, 6690448, 12236830, 21977516, 38754732, 67108864, 114159428, 190876696, 313889477, 508019104, 809785133, 1272196666

Offset: 0

Mokhtar Mohamed, Nov 24 2012

a(n) is the number of compositions (ordered partitions) of n+1 into fourteen or fewer parts.

a(n) is the sum of the first fourteen terms in the n-th row of Pascal's triangle.

			a(14) = 16383 because there are 2^14 = 16384 compositions of 15 into any size parts but one of the compositions (1 + 1 + ... + 1 = 15) has more than fourteen parts.
When 1 <= n <= 13, a(7) = 2*a(6) = 2*64= 128, a(13) = 2*a(12) = 2*4096 = 8192.
When n > 13, a(14) = 2*a(13) - C(13, 13) = 2*8192 - 1 = 16383, a(15) = 2*a(14) - C(14, 13) = 2*16383 - 14 = 32766 - 14 = 32752.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863, A219531.

f:= n -> add(binomial(n,k),k=0..13):
map(f, [$0..100]); # Robert Israel, Mar 14 2018

Table[Sum[Binomial[n, k], {k, 0, 13}], {n, 0, 40}] (* T. D. Noe, Nov 26 2012 *)

a(n) = Sum_{k=1..7} binomial(n+1, 2k-1).
a(n) = 1 +(n^13 -65*n^12 +2015*n^11 -37609*n^10 +470613*n^9 -4081935*n^8 +25378925*n^7 -110205667*n^6 +351042406*n^5 -657328100*n^4 +1303568760*n^3 +771653376*n^2 +4546558080*n)/13!. - corrected by Mokhtar Mohamed, Dec 01 2012
G.f.: (1 - 12*x + 67*x^2 - 230*x^3 + 541*x^4 - 920*x^5 + 1163*x^6 - 1106*x^7 + 791*x^8 - 420*x^9 + 161*x^10 - 42*x^11 + 7*x^12)/(1-x)^14.
a(n) = 2*a(n-1), for 1 <= n <= 13, with a(0) = 1, a(n) = 2*a(n-1) - C(n-1, 13), for n > 13.

Corrected and extended by T. D. Noe, Nov 26 2012

A209257 A musically inspired Titius-Bode-like sequence based on the geometric division of 4- and 5-dimensional space: Z_(n+1) = 3 * (C(n-1, 0) + C(n-1, 1) + C(n-1, 2) + C(n-1, 3) + C(n-1, 4) + C(n-1, 5)*A059620(n+6)) + 4.

Original entry on oeis.org

4, 7, 10, 16, 28, 52, 97, 193, 301, 493, 1150, 1162, 3076, 2386, 3283, 10423, 5827, 20659, 9646, 37852, 15112, 18592, 83692, 27331, 133660, 38857, 45832, 251050, 62566, 367318, 83527, 523315, 109375, 124351, 852826, 158872, 1152508, 200140, 223561, 1754809

Offset: 0

Raphie Frank, Jan 14 2013

The classical Titius-Bode version of this sequence is given in A003461.
C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) + C(n, 4) = A000127(n) = A059173(n+1)/2.
C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) + C(n, 4) + C(n, 5) = A006261(n) = A059174(n+1)/2.
Where planetary and dwarf-planetary distances from the Sun at semi-major axis are expressed in astronomical units/10, then compare the following (noting that the running correlation coefficient, r, trends upwards as the population size increases):
n = 0, Mercury @ semi-major =   3.8710 vs.   4.0 -->  96.78%.
n = 1, Venus   @ semi-major =   7.2333 vs.   7.0 --> 103.33%.
n = 2, Earth   @ semi-major =  10.0000 vs.  10.0 --> 100.00%,  r = 0.998430.
n = 3, Mars    @ semi-major =  15.2368 vs.  16.0 -->  95.23%,  r = 0.998356.
n = 4, Ceres   @ semi-major =  27.654  vs.  28.0 -->  98.76%,  r = 0.999412.
n = 5, Jupiter @ semi-major =  52.0427 vs.  52.0 --> 100.08%,  r = 0.999809.
n = 6, Saturn  @ semi-major =  95.8202 vs.  97.0 -->  98.78%,  r = 0.999937.
n = 7, Uranus  @ semi-major = 192.2941 vs. 193.0 -->  99.63%,  r = 0.999981.
n = 8, Neptune @ semi-major = 301.0366 vs. 301.0 --> 100.01%,  r = 0.999990.
The correspondence between this sequence and planetary distances breaks down subsequent to Neptune unless one adopts the conceit of considering the outer four dwarf planets -- Pluto, Haumea, MakeMake and Eris -- as one unit occupying one "planetary band" (note that Eris @ perihelion is inside the Kuiper Belt). Then:
n = 9, Pluto/Haumea/MakeMake/Eris @ semi-major ~ 490.492 average vs. 493.0 --> 99.49%, r = 0.999994.
Empirical source: Wikipedia planet pages as of Jan 14 2013.
This sequence originated as part of an attempt to compare and contrast the "good" numerology of Johann Balmer to the "bad" numerology of Titius-Bode. Coincidentally, (Totient(C(31, 0) + C(31, 1) + C(31, 2) + C(31, 3) + C(31, 4)))/10^11 equals 3.6456*10^-7, in meters, the Balmer constant as given by Johann Balmer in 1885.

			Z_1 = 3*((1 - 1 +  1 -  1 +  1) + (-1 * 1)) + 4 =   4,
Z_2 = 3*((1 + 0 +  0 +  0 +  0) +  (0 * 0)) + 4 =   7,
Z_3 = 3*((1 + 1 +  0 +  0 +  0) +  (0 * 0)) + 4 =  10,
Z_4 = 3*((1 + 2 +  1 +  0 +  0) +  (0 * 1)) + 4 =  16,
Z_5 = 3*((1 + 3 +  3 +  1 +  0) +  (0 * 0)) + 4 =  28,
Z_6 = 3*((1 + 4 +  6 +  4 +  1) +  (0 * 1)) + 4 =  52,
Z_7 = 3*((1 + 5 + 10 + 10 +  5) +  (1 * 0)) + 4 =  97,
Z_8 = 3*((1 + 6 + 15 + 20 + 15) +  (6 * 1)) + 4 = 193,
Z_9 = 3*((1 + 7 + 21 + 35 + 35) + (21 * 0)) + 4 = 301.

G. C. Greubel, Table of n, a(n) for n = 0..1000
J. Baez, This Week's Finds in Mathematical Physics (Week 234)
H. Chang, Titius-Bode’s Relation and Distribution of Exoplanets
D. M. F. Chapman, The Titius-Bode Rule, Part 2: Science or Numerology?
Chemteam, The Balmer Formula
A. L. Kholodenko, Role of general relativity and quantum mechanics in dynamics of Solar System
Wikipedia, Dwarf planet
Wikipedia, Planet
Wikipedia, Titius-Bode law

Cf. A003461, A059620, A000127, A059173, A006261, A059174.

[3*(Binomial(n-1,0) + Binomial(n-1,1) + Binomial(n-1,2) + Binomial(n-1,3) + Binomial(n-1,4) + Binomial(n-1,5)*(Floor((5*(n+6) + 7)/12) - Floor((5*(n+6)+2)/12))) + 4: n in [0..30]]; // G. C. Greubel, Jan 07 2018

Z[n_]:= 3*(Binomial[n - 1, 0] + Binomial[n - 1, 1] + Binomial[n - 1, 2] + Binomial[n - 1, 3] + Binomial[n - 1, 4] + Binomial[n - 1, 5]*(Floor[(5 (n + 6) + 7)/12] - Floor[(5 (n + 6) + 2)/12])) + 4; Table[Z[n], {n, 0, 50}] (* G. C. Greubel, Jan 07 2018 *)

{z(n) = 3*(binomial(n-1,0) + binomial(n-1,1) + binomial(n-1,2) + binomial(n-1,3) + binomial(n-1,4) + binomial(n-1,5)*(floor((5*(n+6) + 7)/12) - floor((5*(n+6)+2)/12))) + 4};
for(n=0,30, print1(z(n), ", ")) \\ G. C. Greubel, Jan 07 2018

Z_(n+1) = 3 * (C(n-1, 0) + C(n-1, 1) + C(n-1, 2) + C(n-1, 3) + C(n-1, 4) + C(n-1, 5)*(floor((5*(n+6)+7)/12) - floor((5*(n+6)+2)/12))) + 4.

a(18) corrected by G. C. Greubel, Jan 07 2018

A220051 Sum_{k=0..14} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32767, 65519, 130918, 261156, 519252, 1026876, 2014992, 3913704, 7507638, 14198086, 26434916, 48412432, 87167164, 154276028, 268435456, 459312152, 773201629, 1281220733, 2091005866

Offset: 0

Mokhtar Mohamed, Dec 03 2012

a(n) is the number of compositions (ordered partitions) of n+1 into fifteen or fewer parts.
a(n) = sum(binomial(n+1,2k), for k = 0..7).
a(n) is the sum of the first fifteen terms in the n-th row of Pascal's triangle.

			a(15) = 32767 because there are 2^15 = 32768 compositions of 16 into any size parts but one of the compositions (1 + 1 + ... + 1 = 16) has more than fifteen parts.
When 1 <= n <= 14, for n=10, a(10) = 2*a(9) = 2*512 = 1024. For n=14, a(14) = 2*a(13) = 2*8192 = 16384.
When n > 14, for n = 15, a(15) = 2*a(14) -C(14,14) = 2*16384 -1 = 32767. For n=20, a(20) = 2*a(19) -C(19,14) = 2*519252 -11626 = 1038504 -11626 = 1026876.

Indranil Ghosh, Table of n, a(n) for n = 0..1000

Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863, A219531, A219615, A219676.

Table[Sum[Binomial[n,k],{k,0,14}],{n,0,33}] (* Indranil Ghosh, Feb 22 2017 *)
NestList[{#1 + 1, 2 #2 - Boole[#1 >= 14] Binomial[#1, 14]} & @@ # &, {0, 1}, 33][[All, -1]] (* Michael De Vlieger, Feb 22 2017 *)

a(n)=sum(k=0,14,binomial(n,k)) \\ Indranil Ghosh, Feb 23 2017

a(n) = 1 + (n^14 - 77*n^13 + 2821*n^12 - 6288*n^11 + 947947*n^10 - 10081071*n^9 + 77889383*n^8 - 435638203*n^7 + 1793239448*n^6 - 5043110072*n^5 + 1111159696*n^4 - 8346754416*n^3 + 30605906304*n^2 + 57424792320*n)/14!.
G.f.: (1 - 13x + 79x^2 - 297x^3 + 771x^4 - 1461x^5 + 2083x^6 - 2269x^7 + 1897x^8 - 1211x^9 + 581x^10 - 203x^11 + 49x^12 - 7x^13 + x^14)/(1-x)^15.
a(n) = 2*a(n-1), for 1 <= n <= 14, with a(0) = 1, a(n) = 2*a(n-1) - C(n-1,14), for n> 14.

A373005 Array read by ascending antidiagonals: A(n,k) is the maximum possible cardinality of a set of points of diameter at most k-1 in {0,1}^n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 0, 1, 2, 2, 0, 0, 1, 2, 3, 2, 1, 0, 1, 2, 4, 4, 2, 2, 0, 1, 2, 5, 6, 4, 2, 1, 0, 1, 2, 6, 8, 7, 4, 2, 0, 0, 1, 2, 7, 10, 11, 8, 4, 2, 1, 0, 1, 2, 8, 12, 16, 14, 8, 4, 2, 2, 0, 1, 2, 9, 14, 22, 22, 15, 8, 4, 2, 1, 0, 1, 2, 10, 16, 29, 32, 26, 16, 8, 4, 2, 0

Offset: 0

Stefano Spezia, May 19 2024

A(n,k) is also the size of the Hamming ball in {0,1}^n of radius (k-1)/2 if k is odd and of the union of two Hamming balls in {0,1}^n of radius k/2-1 whose centers are of Hamming distance 1 if k is even.

			The array begins:
  1, 1, 2, 1,  0,  1,  2,  1, ...
  0, 1, 2, 2,  2,  2,  2,  2, ...
  0, 1, 2, 3,  4,  4,  4,  4, ...
  0, 1, 2, 4,  6,  7,  8,  8, ...
  0, 1, 2, 5,  8, 11, 14, 15, ...
  0, 1, 2, 6, 10, 16, 22, 26, ...
  0, 1, 2, 7, 12, 22, 32, 42, ...
  0, 1, 2, 8, 14, 29, 44, 64, ...
  ...

Noga Alon, Zhihan Jin, and Benny Sudakov, The Helly number of Hamming balls and related problems, arXiv:2405.10275 [math.CO], 2024. See p. 3.
S. L. Bezrukov, Specification of all maximal subsets of the unit cube with respect to given diameter. Problemy Peredachi Informatsii, pages 106-109, 1987. On ResearchGate.
G. Katona, Intersection theorems for systems of finite sets, Acta Mathematica Academiae Scientiarum Hungaricae 15, 329-337 (1964).
Daniel J. Kleitman, On a combinatorial conjecture of Erdös, Journal of Combinatorial Theory, Series A 1, 209-214, (1966).

Cf. A000007 (k=0), A000012 (k=1), A000124 (k=5), A000125 (k=7), A005843 (k=4), A006261 (k=11), A007395 (k=2), A008859 (k=13), A011782 (main diagonal), A014206, A046127 (k=8), A059173, A059174, A130130 (n=1), A158411 (n=2), A373006 (antidiagonal sums).

A[n_,k_]:=If[OddQ[k],Sum[Binomial[n,i],{i,0,(k-1)/2}], Binomial[n-1,k/2-1]+Sum[Binomial[n,i],{i,0,k/2-1}]]; Table[A[n-k,k],{n,0,12},{k,0,n}]//Flatten

A(n,k) = Sum_{i=0..(k-1)/2} binomial(n,i) if k is odd;
A(n,k) = binomial(n-1,k/2-1) + Sum_{i=0..k/2-1} binomial(n,i) if k is even.
A(n,3) = n+1.
A(n,6) = A014206(n-1).
A(n,9) = A000127(n+1).
A(n,10) = A059173(n) for n > 0.
A(n,12) = A059174(n) for n > 0.
A(0,k) = A007877(k) for k > 0.

cp's OEIS Frontend

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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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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A374378 Iterated rascal triangle R2: T(n,k) = Sum_{m=0..2} binomial(n-k,m)*binomial(k,m).

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A219615 a(n) = Sum_{k=0..12} binomial(n,k).

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A219676 a(n) = Sum_{k=0..13} binomial(n, k).

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Maple

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A209257 A musically inspired Titius-Bode-like sequence based on the geometric division of 4- and 5-dimensional space: Z_(n+1) = 3 * (C(n-1, 0) + C(n-1, 1) + C(n-1, 2) + C(n-1, 3) + C(n-1, 4) + C(n-1, 5)*A059620(n+6)) + 4.

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Magma

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A220051 Sum_{k=0..14} binomial(n,k).

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