A379121 Odd squares k for which A379113(k) > 1, i.e., k that have a proper unitary divisor d > 1 such that A048720(A065621(sigma(d)),sigma(k/d)) is equal to sigma(k).
225, 3025, 3249, 12321, 29241, 38025, 91809, 216225, 247009, 354025, 408321, 751689, 772641, 855625, 919681, 1366561, 1595169, 3814209, 9828225, 11189025, 12173121, 12709225, 29430625, 47927929, 52403121, 66471409, 67486225, 77457601, 80263681, 94148209, 100661089, 110397049, 126540001, 204232681, 264875625, 328878225
Offset: 1
Keywords
Examples
k = 225 = 15^2 is included, because x = A379113(k) = 9, y = A379119(k) = 225/9 = 25, and A048720(A065621(sigma(9)), sigma(25)) = A048720(A065621(13), 31) = A048720(21, 31) = 403 = sigma(225). a(8) = k = 216225 = 465^2 = (3*5*31)^2 is included, because x = A379113(k) = 9, y = A379119(k) = k/9 = 24025, sigma(9) = 13, A065621(13) = 21, sigma(24025) = 30783 and A048720(21, 30783) = 400179 = sigma(k). Note that pair x = 31^2 = 961, y = k / 961 = 225 is not among the solutions (we have A379129(k) = 1, not 2), because A048720(A065621(sigma(961)), sigma(k/961)) = 425971 > 400179. a(520) = k = 383942431613601 = 19594449^2 is included, because x = A379113(k) = 16129, y = A379119(k) = 23804478369, and A048720(A065621(sigma(x)),sigma(y)) = 703777973774337 = sigma(k). This is the first term that has more than one such solution (A379129(k) = 2), the other solution pair being x=961 and y=399523862241. a(1087) = k = 19012955210325729 = 137887473^2 is included, because x = A379113(k) = 8649, y = k/8649 = 2198283640921, and A048720(A065621(sigma(x)),sigma(y)) = A048720(22197, 2198285123583) = sigma(x)*sigma(y) = 28377662660332947 = A379125(1087). Note that 8649 = 9*961 and here also x=961 and x=9 satisfy the condition, so there are three solutions in total.
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