A386520 Column sums of the triangle in A386755.
1, 5, 13, 13, 31, 35, 57, 61, 85, 85, 111, 99, 235, 89, 353, 173, 171, 341, 343, 229, 489, 423, 415, 435, 661, 525, 535, 559, 1161, 427, 931, 653, 1201, 787, 941, 885, 1629, 537, 1443, 1839, 1723, 931, 1119, 1525, 2415, 741, 2257, 2327, 1947, 2005, 2767, 1131, 3181, 1055, 3131, 2147
Offset: 1
Keywords
Examples
Triangle whose columns are summed. m/n| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ---------------------------------------------------------------- 1 | 1 2 | 1 3 | 2 1 4 | 2 1 5 | 3 2 1 6 | 3 2 1 7 | 3 2 1 8 | 3 2 1 9 | 4 3 2 1 10 | 4 3 2 1 11 | 5 4 3 2 1 12 | 5 4 3 2 1 13 | 5 4 3 2 1 14 | 5 4 3 2 1 15 | 5 4 3 2 1 16 | 5 4 3 2 1 17 | 6 5 4 3 2 1 18 | 6 5 4 3 2 1 19 | 6 5 4 3 2 1 20 | 6 5 4 3 2 1 ... The completed column for n=5 is definitely fully visible here because in column 6 for n=6 the divisor k=6 already appeared. That means that column 5 cannot have more divisors in it under the last k=5 in row 17 because in that row only k=7 may follow k=6 in theory, but 7 does not divide 5. So, all similarly proven, definitely fully visible completed columns in this sample array are readily summable by sight. E.g. column 5: a(5) = 1 + 5 + 5 + 5 + 5 + 5 + 5 = 31.
Links
- Michel Marcus, Table of n, a(n) for n = 1..500
Programs
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PARI
\\ uses row(n) from A386755 a(n) = my(ok=1, k=1, last=-1, s=0, r); while(ok, r=row(k); if (#r >= n, s+=r[n]); k++; if (#r>=n, if ((last==n) && (r[n]==0), ok = 0, last = r[n]))); s; \\ Michel Marcus, Aug 02 2025
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PARI
\\ uses row(n) from A386755 lista(nn) = my(ok=1, k=1, vlast=vector(nn,i,-1), vs=vector(nn)); while(ok, my(r=row(k)); for (i=1, nn, if (#r>=i, vs[i]+=r[i])); k++; my(nbok=0); for (i=1, nn, if (#r>=i, if ((vlast[i]==i) && (r[i]==0), nbok++, vlast[i] = r[i]))); if (nbok == nn, ok = 0);); vs; \\ Michel Marcus, Aug 02 2025
Comments