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A347839
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An array of the positive integers congruent to 2 modulo 3 (A016789), read by antidiagonals upwards, giving the present triangle.
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3
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2, 5, 8, 11, 20, 32, 14, 44, 80, 128, 17, 56, 176, 320, 512, 23, 68, 224, 704, 1280, 2048, 26, 92, 272, 896, 2816, 5120, 8192, 29, 104, 368, 1088, 3584, 11264, 20480, 32768, 35, 116, 416, 1472, 4352, 14336, 45056, 81920, 131072, 38, 140, 464, 1664, 5888, 17408, 57344, 180224, 327680, 524288
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OFFSET
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1,1
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COMMENTS
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This array a = (a(k, n))_{k >= 1,n >= 0} is underlying array A of A347834. See the first formula. It has a simple recurrence for the rows k, given the first column a(k, 0) = A347838(k), which lists the positive integers congruent to {2, 5, 11} modulo 12.
In the array one can add the negative of the powers of 4 as row for k = 0, i.e., -A000302(n), for n >= 0.
All positive numbers congruent to 2 modulo 3 (A017617) appear once in this array. Proof from the array A of A347834 of the positive integers congruent to {1,3,5,7} modulo 8, and the present first formula: The members of column n = 0 give all the positive integers congruent to {2, 5, 11} modulo 12 once, and the members of columns n >= 1 give all the positive integers congruent to 8 modulo 12 (A017617) once. These members combined lead to the positive integers congruent to 2 modulo 3.
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LINKS
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FORMULA
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Array a:
a(k, n) = (3*A(k, n) + 1)/2, with the array A from A347834, for k >= 1, and n >= 0.
a(k, n) = 4^n*A347838(k) = 4^n*(2 + 3*k + 3*floor((k + 1)/3)).
Recurrence for rows k: a(k, n) = 4*a(k, n-1), for n >= 1, with a(k, 0) = A347838(k).
O.g.f.: expansion in z gives the o.g.f.s for rows k, also for k = 0: -A000302; expansion in x gives the o.g.f.s for columns n.
G(z, x) = (-1 + 3*z + 3*z^2 + 7*z^3)/((1 - z)*(1 - z^3)*(1 - 4*x)).
Triangle t:
t(k, n) = a(k-n, n), for k >= 1, and n = 0, 1, ..., k-1.
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EXAMPLE
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The array a(k, n) begins:
k \ n 0 1 2 3 4 5 6 7 8 9 10 ...
---------------------------------------------------------------------------
1: 2 8 32 128 512 2048 8192 32768 131072 524288 2097152 ...
2: 5 20 80 320 1280 5120 20480 81920 327680 1310720 5242880 ...
3: 11 44 176 704 2816 11264 45056 180224 720896 2883584 11534336 ...
4: 14 56 224 896 3584 14336 57344 229376 917504 3670016 14680064 ...
5: 17 68 272 1088 4352 17408 69632 278528 1114112 4456448 17825792 ...
6: 23 92 368 1472 5888 23552 94208 376832 1507328 6029312 24117248 ...
7: 26 104 416 1664 6656 26624 106496 425984 1703936 6815744 27262976 ...
8: 29 116 464 1856 7424 29696 118784 475136 1900544 7602176 30408704 ...
9: 35 140 560 2240 8960 35840 143360 573440 2293760 9175040 36700160 ...
10: 38 152 608 2432 9728 38912 155648 622592 2490368 9961472 39845888 ...
...
----------------------------------------------------------------------------
The triangle t(n,k) begins:
k \ n 0 1 2 3 4 5 6 7 8 9 ...
---------------------------------------------------------------
1: 2
2: 5 8
3: 11 20 32
4: 14 44 80 128
5: 17 56 176 320 512
6: 23 68 224 704 1280 2048
7: 26 92 272 896 2816 5120 8192
8: 29 104 368 1088 3584 11264 20480 32768
9: 35 116 416 1472 4352 14336 45056 81920 131072
10: 38 140 464 1664 5888 17408 57344 180224 327680 524288
...
-----------------------------------------------------------------
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MAPLE
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A := (n, k) -> 4^n*(3*(k + iquo(k, 3)) - 1):
for k from 1 to 10 do seq(A(n, k), n = 0..10) od;
# Alternatively:
gf := n -> (4^n*((z*(z*(7*z + 3) + 3) - 1)))/((z - 1)^2*(1 + z + z^2)):
ser := n -> series(gf(n), z, 12):
col := (n, len) -> seq(coeff(ser(n), z, k), k = 1..len):
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MATHEMATICA
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A[n_, k_] := 4^n (3(k + Quotient[k, 3]) - 1);
Table[A[n-k, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 07 2021, from Maple code *)
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CROSSREFS
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The columns n are given by 4^n*A347838 for n >= 0.
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KEYWORD
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AUTHOR
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STATUS
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approved
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