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A093915
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Triangle with r-th row containing r consecutive integers that sum to the smallest possible proper multiple of A006003(r).
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4
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2, 7, 8, 9, 10, 11, 24, 25, 26, 27, 24, 25, 26, 27, 28, 53, 54, 55, 56, 57, 58, 47, 48, 49, 50, 51, 52, 53, 94, 95, 96, 97, 98, 99, 100, 101, 78, 79, 80, 81, 82, 83, 84, 85, 86, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 212, 213, 214, 215, 216, 217
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OFFSET
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1,1
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COMMENTS
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The r-th row constructed as explained in the example starts with x=A093916(r), ends with x+r-1=A093918(r), and has its sum A093917(r) equal to the smallest proper multiple of A006003(r). There is a simple formula for A093917(r), which allows us to calculate A093915(n) directly. - M. F. Hasler, Apr 04 2009
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LINKS
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EXAMPLE
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Given the triangle
1 . . . . with row sum S1 = 1 = A006003(1)
2,3 . . . with row sum S2 = 2+3 = 5 = A006003(2)
4,5,6 . . with row sum S3 = 4+5+6 = 15 = A006003(3), etc.,
the sequence is constructed as follows:
The first row below must be a proper (i.e., > 1) multiple of S1; the smallest possibility is [ 2 ].
The next row below must contain 2 consecutive integers with sum equal to a proper multiple of S2=5. It cannot be 10 (not the sum of 2 consecutive integers), but 15 = 7+8 is a possibility.
The third row [x,x+1,x+2] must sum to a multiple of S3=15, and 2*S3=30 is possible for x=9.
The 4th row [x,x+1,x+2,x+3] must have its sum 4x+6 equal to a multiple of S4=7+8+9+10=34, and x=24 gives the sum 102=3*34, while 2*34=68 can't be achieved for any integer x.
This gives:
2 . . . . . . . with row sum 2 = 2*S1
7,8 . . . . . . with row sum 7+8 = 15 = 3*S2
9,10,11 . . . . with row sum 9+10+11 = 30 = 2*S3
24,25,26,27 . . with row sum 24+25+26+27 = 102 = 3*S4.
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PROG
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(PARI) for(r=1, 9, x=A093916(r)-1; for(c=1, r, print1(x+c, ", "))) /* M. F. Hasler, Apr 04 2009 */
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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Edited and extended (values beyond a(15), example, PARI code) by M. F. Hasler, Apr 04 2009
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STATUS
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approved
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