[seqfan] Re: Sums of n products of pairs of 0..k integers?

Ron Hardin rhhardin at att.net
Tue Apr 16 17:19:08 CEST 2013


A sum of n products of 3, instead of 2, 0..k integers behaves similarly

/tmp/dgv
T(n,k)=Number of distinct values of the sum of n products of three 0..k integers

Table starts
..2..5..11..17...31...41...66...81..101..121...174...195...267...302...344
..3.12..36..75..157..254..434..635..911.1237..1734..2162..2908..3611..4461
..4.20..63.141..284..478..780.1156.1667.2298..3111..4012..5213..6520..8063
..5.28..90.205..409..694.1123.1668.2396.3298..4442..5741..7410..9266.11443
..6.36.117.269..534..910.1466.2180.3125.4298..5773..7469..9607.12010.14818
..7.44.144.333..659.1126.1809.2692.3854.5298..7104..9197.11804.14754.18193
..8.52.171.397..784.1342.2152.3204.4583.6298..8435.10925.14001.17498.21568
..9.60.198.461..909.1558.2495.3716.5312.7298..9766.12653.16198.20242.24943
.10.68.225.525.1034.1774.2838.4228.6041.8298.11097.14381.18395.22986.28318
.11.76.252.589.1159.1990.3181.4740.6770.9298.12428.16109.20592.25730.31693

Row 1 is A027426


T(n,1) = 1*n + 1 for n>0
T(n,2) = 8*n - 4 for n>1
T(n,3) = 27*n - 18 for n>1
T(n,4) = 64*n - 51 for n>2
T(n,5) = 125*n - 91 for n>2
T(n,6) = 216*n - 170 for n>2
T(n,7) = 343*n - 249 for n>2
T(n,8) = 512*n - 380 for n>2
T(n,9) = 729*n - 520 for n>2
T(n,10) = 1000*n - 702 for n>2
T(n,11) = 1331*n - 882 for n>2
T(n,12) = 1728*n - 1171 for n>3
T(n,13) = 2197*n - 1378 for n>2  <--- curious
T(n,14) = 2744*n - 1710 for n>3
T(n,15) = 3375*n - 2057 for n>3
... >3's ...
T(n,55) = 166375*n - 52512 for n>3


 rhhardin at mindspring.com
rhhardin at att.net (either)


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