A053152 -id:A053152 - cp's OEIS frontend

A250755 T(n,k)=Number of (n+1)X(k+1) 0..2 arrays with nondecreasing x(i,j)-x(i,j-1) in the i direction and nondecreasing x(i,j)+x(i-1,j) in the j direction.

32, 72, 105, 129, 237, 332, 203, 423, 756, 1029, 294, 663, 1353, 2361, 3152, 402, 957, 2123, 4239, 7272, 9585, 527, 1305, 3066, 6663, 13089, 22197, 29012, 669, 1707, 4182, 9633, 20603, 40023, 67356, 87549, 828, 2163, 5471, 13149, 29814, 63063, 121593

Offset: 1

R. H. Hardin, Nov 27 2014

Table starts
.....32......72.....129.....203.....294......402......527......669......828
....105.....237.....423.....663.....957.....1305.....1707.....2163.....2673
....332.....756....1353....2123....3066.....4182.....5471.....6933.....8568
...1029....2361....4239....6663....9633....13149....17211....21819....26973
...3152....7272...13089...20603...29814....40722....53327....67629....83628
...9585...22197...40023...63063...91317...124785...163467...207363...256473
..29012...67356..121593..191723..277746...379662...497471...631173...780768
..87549..203601..367839..580263..840873..1149669..1506651..1911819..2365173
.263672..613872.1109649.1751003.2537934..3470442..4548527..5772189..7141428
.793065.1847757.3341223.5273463.7644477.10454265.13702827.17390163.21516273

			Some solutions for n=4 k=4
..1..1..1..1..0....0..0..0..0..0....0..0..0..0..0....0..0..0..0..0
..1..1..1..1..2....2..2..2..2..2....1..1..1..1..1....0..0..0..0..0
..1..1..1..1..2....0..0..0..0..0....0..0..0..0..0....1..1..1..1..1
..0..1..1..1..2....1..1..1..2..2....2..2..2..2..2....0..0..2..2..2
..0..1..1..1..2....0..0..0..1..2....0..1..1..1..2....0..0..2..2..2

R. H. Hardin, Table of n, a(n) for n = 1..241

Column 1 is A053152(n+3)

Empirical: T(n,k) = (3*(k+1)*(5*k+4)*3^n - (8*k^2+8*k)*2^n + (5*k^2-7*k))/4
Empirical for column k:
k=1: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (27*3^n-8*2^n-1)/2
k=2: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (63*3^n-24*2^n+3)/2
k=3: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (114*3^n-48*2^n+12)/2
k=4: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (180*3^n-80*2^n+26)/2
k=5: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (261*3^n-120*2^n+45)/2
k=6: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (357*3^n-168*2^n+69)/2
k=7: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (468*3^n-224*2^n+98)/2
Empirical for row n:
n=1: a(n) = (17/2)*n^2 + (29/2)*n + 9
n=2: a(n) = 27*n^2 + 51*n + 27
n=3: a(n) = (173/2)*n^2 + (329/2)*n + 81
n=4: a(n) = 273*n^2 + 513*n + 243
n=5: a(n) = (1697/2)*n^2 + (3149/2)*n + 729
n=6: a(n) = 2607*n^2 + 4791*n + 2187
n=7: a(n) = (15893/2)*n^2 + (29009/2)*n + 6561

A064686 a(n) = number of n-digit base-3 biquams.

Original entry on oeis.org

0, 2, 7, 23, 73, 227, 697, 2123, 6433, 19427, 58537, 176123, 529393, 1590227, 4774777, 14332523, 43013953, 129074627, 387289417, 1161999323, 3486260113, 10459304627, 31378962457, 94138984523, 282421147873, 847271832227

Offset: 1

David W. Wilson, Oct 10 2001

A biquam or biquanimous number (A064544) is a number whose digits can be split into two groups with equal sum.
This is the same as A083313 (apart from the initial term). Proof: Let sum(w) denote the sum of the digits of w. There are 2*3^(n-1) n-digit base-3 numbers: w = (w_1,w_2,...,w_n) with w_i in {0,1,2} for all i and w_1 != 0. Partition them into 4 classes: (i) sum(w) is odd, (ii) sum(w) is even, w contains no 1's and has an odd number of 2s, (iii) sum(w) is even, w contains no 1's and has an even number of 2s and (iv) sum(w) is even and w contains some 1's. Clearly, no biquams occur in cases (i) and (ii), case (iii) consists entirely of biquams and, we claim, so does case (iv). For case (iv) forces an even number, say 2k, of 1's. An even number of 2s clearly gives a biquam and an odd number 2m+1 of 2s does too because {m 2s, (k+1) 1's} and {(m+1) 2s, (k-1) 1's} is a biquam split. There are 3^(n-1) w's in case (i) and 2^(n-2) w's in case (ii) and hence 2*3^(n-1) - (3^(n-1) + 2^(n-2)) = 3^(n-1) - 2^(n-2) (A083313) biquams among n-digit base-3 numbers. - David Callan, Sep 15 2004
a(n) % 100 = 23  for n = 4*k-1, k>=1; a(n) % 100 = 27  for n = 4*k+1, k>=1. - Alex Ratushnyak, Jul 03 2012
The fraction of biquams for any base approaches 1/2 as the number of digits grows but only if you count leading zeros. Without counting leading zeros, the fraction appears to converge to (b-1)/2b where b is the base used. For base 3 this is 1/3 which fits the data in this sequence (see paper cited below for proofs and the OEIS data collated as fractions). - Timothy Varghese, Aug 08 2021

Timothy Varghese and George Varghese, The Stash-Repair Method applied to Partitionable Numbers,
Index entries for linear recurrences with constant coefficients, signature (5,-6).

Essentially the same as A083313.

Cf. A053152 (partial sums).

print([0]+[3**n - 2**(n-1) for n in range(1,29)])
# Alex Ratushnyak, Jul 02 2012

a(1) = 0, a(n) = 3^(n-1)-2^(n-2) for n>=2. - Alex Ratushnyak, Jul 02 2012

a(n) = 5*a(n-1)-6*a(n-2) for n>3. G.f.: -x^2*(3*x-2) / ((2*x-1)*(3*x-1)). - Colin Barker, May 27 2013

A191487 The row sums of the Sierpinski-Stern triangle A191372.

Original entry on oeis.org

0, 1, 3, 8, 9, 22, 24, 26, 27, 62, 66, 70, 72, 76, 78, 80, 81, 178, 186, 194, 198, 206, 210, 214, 216, 224, 228, 232, 234, 238, 240, 242, 243, 518, 534, 550, 558, 574, 582, 590, 594, 610, 618, 626, 630, 638, 642, 646

Offset: 0

Johannes W. Meijer, Jun 05 2011

nonn

The row sums a(n) of the Sierpinski-Stern triangle A191372 equal this sequence.
The differences diff1(n) = a(2*n+3) - a(2*n+1) and diff2(n) = (a(2*n+2) - a(2*n))/3, give rise to patterns that lead to Gould’s sequence A001316, see the examples.
The diff1(n) sequence as a triangle leads to Gould’s sequence in a peculiar way, see A191488. The leading terms of the diff1(n) rows are given by A001550(p+1), p>=1; for p=0 the leading term is 7. The rows sums of diff1(n) as a triangle equal A025192(p+2), p>=1; for p = 0 the row sum is 7. The row sums of diff1(n) as a triangle minus the first term equal 2*A053152(p+1).
The diff2(n) sequence as a triangle leads to Gould’s sequence A001316 in a simple way; just delete the first term and reverse the order of the rest of the terms; more terms require a higher row number. The leading terms of the diff2(n) rows are given by A085281(p), p>=0. The row sums of diff2(n) as a triangle equal A025192(p) and the row sums minus the first term equal A001047(p-1), p>=1; for p=0 the row sum minus the first term is 0.

			The first few rows of diff1(n) as a triangle, row lengths A000079(p) with p>=0, are:
[7]
[14, 4]
[36, 8, 6, 4]
[98, 16, 12, 8, 10, 8, 6, 4]
[276, 32, 24, 16, 20, 16, 12, 8, 18, 16, 12, 8, 10, 8, 6, 4]
[794, 64, 48, 32, 40, 32, 24, 16, 36, 32, 24, 16, 20, 16, 12, 8, 34, 32, 24, 16, 20, 16, 12, 8, 18, 16, 12, 8, 10, 8, 6, 4]
The first few rows of diff2(n) as a triangle, row lengths A011782(p) with p>=0, are:
[1]
[2]
[5, 1]
[13, 2, 2, 1]
[35, 4, 4, 2, 4, 2, 2, 1]
[97, 8, 8, 4, 8, 4, 4, 2, 8, 4, 4, 2, 4, 2, 2, 1]
[275, 16, 16, 8, 16, 8, 8, 4, 16, 8, 8, 4, 8, 4, 4, 2, 16, 8, 8, 4, 8, 4, 4, 2, 8, 4, 4, 2, 4, 2, 2, 1]

Cf. A191372, A191488, A001316, A001550, A025192, A053152, A085281, A001047, A007689.

Add the following lines to the Maple program of A191372.
A191487(0):=0: for d from 1 to 2^pmax do A191487(d):= 0: for Tx from 0 to 2^ceil(log(d)/ log(2))-1 do A191487(d):=A191487(d)+S2(Tx,d) od: od: seq(A191487(d),d=0..2^pmax);

a(2*n) = 3*a(n)
diff(n) = a(n+1) - a(n), diff1(n) = a(2*n+3) - a(2*n+1), diff2(n) = (a(2*n+2) - a(2*n))/3
a(2^n+1) - a(2^n) = A085281(n+1) = A007689(n) for n>=0
a(2^(n+1)+1) - a(2^(n+1)-1) = A001550(n+1) for n>=1.

A052391 Number of 4-element intersecting families (of distinct sets) whose union is an n-element set.

Original entry on oeis.org

0, 0, 4, 349, 9985, 213230, 4000444, 69940479, 1170549895, 19024433560, 302846958634, 4748624978009, 73628721516805, 1132119741733890, 17298702716660824, 263082403948681939, 3986935934969727715

Offset: 1

Vladeta Jovovic, Goran Kilibarda, Mar 11 2000

G. C. Greubel, Table of n, a(n) for n = 1..845
V. Jovovic, G. Kilibarda, On the number of Boolean functions in the Post classes F^{mu}_8, Diskretnaya Matematika, 11 (1999), no. 4, 127-138.
V. Jovovic, G. Kilibarda, On the number of Boolean functions in the Post classes F^{mu}_8, (English translation), Discrete Mathematics and Applications, 9, (1999), no. 6.
Index entries for linear recurrences with constant coefficients, signature (71, -2205, 39495, -452523, 3473673, -18166175, 64427005, -150923976, 220549356, -178819920, 59875200).

Cf. A051181, A053152, A053153.

[(15^n - 6*11^n + 12*9^n - 8^n - 22*7^n + 15*6^n + 12*5^n - 17*4^n + 17*3^n - 11*2^n - 6)/24: n in [0..50]]; // G. C. Greubel, Oct 08 2017

Table[(15^n - 6*11^n + 12*9^n - 8^n - 22*7^n + 15*6^n + 12*5^n - 17*4^n + 17*3^n - 11*2^n - 6)/4!, {n, 0, 50}] (* G. C. Greubel, Oct 08 2017 *)
LinearRecurrence[{71,-2205,39495,-452523,3473673,-18166175,64427005,-150923976,220549356,-178819920,59875200},{0,0,4,349,9985,213230,4000444,69940479,1170549895,19024433560,302846958634},20] (* Harvey P. Dale, May 20 2018 *)

for(n=0,50, print1((15^n - 6*11^n + 12*9^n - 8^n - 22*7^n + 15*6^n + 12*5^n - 17*4^n + 17*3^n - 11*2^n - 6)/24, ", ")) \\ G. C. Greubel, Oct 08 2017

a(n) = (15^n - 6*11^n + 12*9^n - 8^n - 22*7^n + 15*6^n + 12*5^n - 17*4^n + 17*3^n - 11*2^n - 6)/4!.

G.f.: x^3*(14968800*x^8 - 25752870*x^7 + 16968966*x^6 - 5759365*x^5 + 1095624*x^4 - 115860*x^3 + 5974*x^2 - 65*x - 4)/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)*(9*x-1)*(11*x-1)*(15*x-1)). - Colin Barker, Jul 30 2012

A362316 Expansion of e.g.f (exp(x)-1)(exp(2x)-1).

Original entry on oeis.org

0, 0, 4, 18, 64, 210, 664, 2058, 6304, 19170, 58024, 175098, 527344, 1586130, 4766584, 14316138, 42981184, 129009090, 387158344, 1161737178, 3485735824, 10458256050, 31376865304, 94134790218, 282412759264, 847255055010, 2541798719464, 7625463267258, 22876524019504, 68629840493970

Offset: 0

Enrique Navarrete, Apr 16 2023

Number of ternary strings with at least one 0 or one 1 and at least one 2.

			The 4 strings for n=4 are 12, 21, 02, 20.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-11,6).

Cf. A001550, A053152, A083321 (alternates signs).

LinearRecurrence[{6,-11,6},{0,0,4,18},30] (* Paolo Xausa, Aug 07 2023 *)

a(n) = 3^n - 2^n - 1, n>0; a(0)=0.

a(n) = 2*A053152(n).

cp's OEIS Frontend

A250755 T(n,k)=Number of (n+1)X(k+1) 0..2 arrays with nondecreasing x(i,j)-x(i,j-1) in the i direction and nondecreasing x(i,j)+x(i-1,j) in the j direction.

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A064686 a(n) = number of n-digit base-3 biquams.

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Python

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A191487 The row sums of the Sierpinski-Stern triangle A191372.

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A052391 Number of 4-element intersecting families (of distinct sets) whose union is an n-element set.

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Magma

Mathematica

PARI

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A362316 Expansion of e.g.f (exp(x)-1)*(exp(2*x)-1).

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Mathematica

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A362316 Expansion of e.g.f (exp(x)-1)(exp(2x)-1).