A353149 Sum of the odd-indexed terms in the n-th row of the triangle A196020.
1, 3, 5, 7, 9, 12, 13, 15, 20, 19, 21, 28, 25, 27, 37, 31, 33, 44, 37, 42, 52, 43, 45, 60, 54, 51, 68, 56, 57, 83, 61, 63, 84, 67, 81, 92, 73, 75, 100, 90, 81, 113, 85, 87, 130, 91, 93, 124, 104, 114, 132, 103, 105, 143, 126, 120, 148, 115, 117, 175, 121, 123, 180, 127, 150, 173, 133, 135, 180, 175
Offset: 1
Keywords
Examples
For n = 15 the 15th row of the triangle A196020 is [29, 13, 7, 0, 1]. The sum of the odd-indexed terms is 29 + 7 + 1 = 37, so a(15) = 37. Illustration of a(15) = 37: Level Diagram . _ 1 _|1|_ 2 _|1 1|_ 3 _|1 1|_ 4 _|1 1|_ 5 _|1 _ 1|_ 6 _|1 |1| 1|_ 7 _|1 | | 1|_ 8 _|1 _| |_ 1|_ 9 _|1 |1 1| 1|_ 10 _|1 | | 1|_ 11 _|1 _| |_ 1|_ 12 _|1 |1 1| 1|_ 13 _|1 | | 1|_ 14 _|1 _| _ |_ 1|_ 15 |1 |1 |1| 1| 1| . The diagram has 37 steps, so a(15) = 37.
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000
- OEIS Plot 2, A353149 vs A000203
Crossrefs
Programs
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PARI
a(n) = { my(r = A196020row(n)); sum(i = 0, (#r-1)\2, r[2*i + 1]) } A196020row(n) = { my(res, qc); qc = (sqrtint(8*n + 1) - 1)\2; res = vector(qc); for(i = 1, qc, cn = n - binomial(i + 1, 2); if(cn % i == 0, res[i] = 2*(cn/i) + 1 ) ); res } \\ David A. Corneth, Apr 28 2022
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