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A252117 Triangle read by row: T(n,k), n>=1, k>=1, in which column k lists the numbers of A000716 multiplied by A000330(k), and the first element of column k is in row A000217(k). 1
1, 3, 9, 5, 22, 15, 51, 45, 108, 110, 14, 221, 255, 42, 429, 540, 126, 810, 1105, 308, 1479, 2145, 714, 30, 2640, 4050, 1512, 90, 4599, 7395, 3094, 270, 7868, 13200, 6006, 660, 13209, 22995, 11340, 1530, 21843, 39340, 20706, 3240, 55, 35581, 66045, 36960, 6630, 165, 57222, 109215, 64386, 12870, 495 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Conjecture: gives an identity for sigma. Alternating sum of row n equals the sum of divisors of n, i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A000203(n).

Row n has length A003056(n) hence column k starts in row A000217(k).

Column 1 is A000716, but here the offset is 1 not 0.

The 1st element of column k is A000330(k).

The 2nd element of column k is A059270(k).

The 3rd element of column k is A220443(k).

The partial sums of column k give the k-th column of A249120.

This triangle has been constructed after Mircea Merca's formula for A000203.

LINKS

Table of n, a(n) for n=1..55.

EXAMPLE

Triangle begins:

1;

3,

9,           5;

22,         15,

51,         45,

108,       110,     14;

221,       255,     42,

429,       540,    126,

810,      1105,    308,

1479,     2145,    714,     30;

2640,     4050,   1512,     90,

4599,     7395,   3094,    270,

7868,    13200,   6006,    660,

13209,   22995,  11340,   1530,

21843,   39340,  20706,   3240,    55;

35581,   66045,  36960,   6630,   165,

57222,  109215,  64386,  12870,   495,

90882,  177905, 110152,  24300,  1210,

142769, 286110, 184926,  44370,  2805,

221910, 454410, 305802,  79200,  5940,

341649, 713845, 498134, 137970, 12155,    91;

...

For n = 6 the divisors of 6 are 1, 2, 3, 6, so the sum of divisors of 6 is 1 + 2 + 3 + 6 = 12. On the other hand, the 6th row of the triangle is 108, 110, 14, so the alternating row sum is 108 - 110 + 14 = 12, equaling the sum of divisors of 6.

For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 21843, 39340, 20706, 3240, 55, so the alternating row sum is 21843 - 39340 + 20706 - 3240 + 55 = 24, equaling the sum of divisors of 15.

CROSSREFS

Cf. A000203, A000217, A000330, A000716, A003056, A059270, A196020, A220443, A238442, A249120.

Sequence in context: A088898 A143218 A262024 * A103934 A186814 A077384

Adjacent sequences:  A252114 A252115 A252116 * A252118 A252119 A252120

KEYWORD

nonn,tabf

AUTHOR

Omar E. Pol, Dec 14 2014

STATUS

approved

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Last modified February 27 15:59 EST 2020. Contains 332307 sequences. (Running on oeis4.)